AtCoder Educational DP Contest
A
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int n, k;
int f[N], a[N];
int main() {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
for(int i = 2; i <= n; ++i) {
f[i] = 0x3f3f3f3f;
if(i >= 2) f[i] = min(f[i], f[i - 1] + abs(a[i] - a[i - 1]));
if(i >= 3) f[i] = min(f[i], f[i - 2] + abs(a[i] - a[i - 2]));
}
printf("%d\n", f[n]);
}
B
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int n, k;
int f[N], a[N];
int main() {
scanf("%d%d", &n, &k);
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
for(int i = 2; i <= n; ++i) {
f[i] = 0x3f3f3f3f;
for(int j = max(i - k, 1); j < i; ++j) {
f[i] = min(f[i], f[j] + abs(a[i] - a[j]));
}
}
printf("%d\n", f[n]);
}
C
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int n, k;
int f[N][3], a[N][3];
int main() {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) scanf("%d%d%d", &a[i][0], &a[i][1], &a[i][2]);
for(int i = 1; i <= n; ++i) {
for(int j = 0; j < 3; ++j) {
for(int k = 0; k < 3; ++k) {
if(j == k) continue;
f[i][j] = max(f[i][j], f[i - 1][k] + a[i][j]);
}
}
}
printf("%d\n", max(max(f[n][0], f[n][1]), f[n][2]));
}
D
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 100010;
int n, W;
ll f[N], w[N], v[N];
int main() {
scanf("%d%d", &n, &W);
ll ans = 0;
for(int i = 1; i <= n; ++i) scanf("%lld%lld", &w[i], &v[i]);
for(int i = 1; i <= n; ++i) {
for(int j = W; j >= w[i]; --j) {
f[j] = max(f[j], f[j - w[i]] + v[i]);
}
}
for(int i = 1; i <= W; ++i) ans = max(ans, f[i]);
printf("%lld\n", ans);
}
E
對比前面四題不那麼傻的一題
很妙的一個轉化，注意到這題的\(W\)很大，可是\(v\)很小，因此不妨轉換一下狀態，設\(f[i,j]\)表示前\(i\)個物品，選出後的總價值爲\(j\)，須要的最小體積。那麼有\(f[i][j]=\min\{f[i-1][j-v[i]]+w[i]\}\)，答案就是對於全部\(f[n][j]\le W\)的\(j\)取\(\max\)。c++
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 100010;
int n, v[N], W, w[N];
int f[110][N];
int main() {
scanf("%d%d", &n, &W);
int sum = 0;
for(int i = 1; i <= n; ++i) scanf("%d%d", &w[i], &v[i]), sum += v[i];
memset(f, 0x3f, sizeof(f));
f[0][0] = 0;
int ans = 0;
for(int i = 1; i <= n; ++i) {
for(int j = 0; j <= sum; ++j) {
if(j >= v[i]) f[i][j] = min(f[i][j], f[i - 1][j - v[i]] + w[i]);
f[i][j] = min(f[i][j], f[i - 1][j]);
if(f[i][j] <= W) ans = max(ans, j);
}
}
printf("%d\n", ans);
}
F
LCS作一下就好，輸出方案的話從\([n,m]\)往回找就行了，用個棧壓一下答案。函數
#include <bits/stdc++.h>
using namespace std;
const int N = 3010;
char s[N], t[N], st[N];
int f[N][N];
int main() {
scanf("%s%s", s + 1, t + 1);
int n = strlen(s + 1), m = strlen(t + 1);
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= m; ++j) {
f[i][j] = max(f[i - 1][j], f[i][j - 1]);
if(s[i] == t[j]) f[i][j] = max(f[i][j], f[i - 1][j - 1] + 1);
}
}
int now = f[n][m] + 1;
int top = 0;
for(int i = n; i; --i) {
for(int j = m; j; --j) {
if(s[i] == t[j] && f[i][j] == now - 1) {
--now;
st[++top] = s[i];
break;
}
}
}
while(top) putchar(st[top--]);
}
G
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int n, m, deg[N], q[N], f[N];
int cnt, head[N];
struct edge { int to, nxt; } e[N];
void ins(int u, int v) {
e[++cnt] = (edge) { v, head[u] };
head[u] = cnt;
}
int main() {
scanf("%d%d", &n, &m);
for(int i = 1, u, v; i <= m; ++i) {
scanf("%d%d", &u, &v);
ins(u, v); deg[v]++;
}
int l = 1, r = 1;
for(int i = 1; i <= n; ++i) {
if(!deg[i]) q[r++] = i;
}
while(l < r) {
int u = q[l++];
for(int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
f[v] = max(f[v], f[u] + 1);
deg[v]--;
if(!deg[v]) q[r++] = v;
}
}
int ans = 0;
for(int i = 1; i <= n ;++i) ans = max(ans, f[i]);
printf("%d\n", ans);
return 0;
}
H
#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
const int mod = 1e9 + 7;
int h, w, f[N][N];
char a[N][N];
int main() {
scanf("%d%d", &h, &w);
f[1][1] = 1;
for(int i = 1; i <= h; ++i) scanf("%s", a[i] + 1);
for(int i = 1; i <= h; ++i) {
for(int j = 1; j <= w; ++j) {
if(i == 1 && j == 1) continue;
if(a[i][j] == '.' && a[i - 1][j] == '.')
(f[i][j] += f[i - 1][j]) %= mod;
if(a[i][j] == '.' && a[i][j - 1] == '.')
(f[i][j] += f[i][j - 1]) %= mod;
}
}
printf("%d\n", f[h][w]);
}
I
#include <bits/stdc++.h>
using namespace std;
const int N = 3000;
int n;
double p[N], f[2][N];
int main() {
scanf("%d", &n);
int cur = 0;
for(int i = 1; i <= n; ++i) scanf("%lf", &p[i]);
f[1][0] = 1;
for(int i = 1; i <= n; ++i) {
memset(f[cur], 0, sizeof(f[cur]));
for(int j = 0; j <= i; ++j) {
f[cur][j] += f[cur ^ 1][j] * (1 - p[i]);
f[cur][j] += f[cur ^ 1][j - 1] * p[i];
}
cur ^= 1;
}
double ans = 0;
for(int i = n / 2 + 1; i <= n; ++i) ans += f[cur ^ 1][i];
printf("%.15lf\n", ans);
}
J
設\(f[i][j][k]\)表示當前\(a_x=3\)的數有\(i\)個，\(=2\)的有\(j\)個，\(=1\)的有\(k\)個。
則有方程
\[ f[i][j][k]=\frac {\left( 1+f[i-1][j+1][k]*\frac i n+f[i][j-1][k+1] * \frac j n + f[i][j][k-1]* \frac k n \right)}{\frac {i+j+k} n} \]
答案即爲\(f[cnt[3]][cnt[2]][cnt[1]]\)優化
#include <bits/stdc++.h>
using namespace std;
const int N = 310;
double f[N][N][N];
int n, a[N], cnt[5];
int main() {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]), cnt[a[i]]++;
for(int i = 0; i <= n; ++i) {
for(int j = 0; j <= n; ++j) {
for(int k = 0; k <= n; ++k) {
if(i + j + k > n) continue;
if(!i && !j && !k) continue;
if(i > cnt[3]) continue;
if(j > cnt[3] + cnt[2]) continue;
if(k > cnt[3] + cnt[2] + cnt[1]) continue;
double p1 = 1.0 * i / n, p2 = 1.0 * j / n, p3 = 1.0 * k / n;
double p4 = 1.0 * (i + j + k) / n;
f[i][j][k] = (double)(1+(i?f[i-1][j+1][k]:0)*p1+(j?f[i][j-1][k+1]:0)*p2+(k?f[i][j][k-1]:0)*p3)/p4;
}
}
}
printf("%.10lf\n", f[cnt[3]][cnt[2]][cnt[1]]);
}
K
\(SG\)函數板子題，根據\(SG\)定理，只須要\(sg(k)\)不爲\(0\)就先手必勝。
對於\(\text{mex}\)運算我直接從第一個數開始枚舉了...須要複雜度正確的話就須要寫個主席樹或者寫個權值分塊。複雜度是\(O(nk\log A)\)或者\(O(nk \sqrt A)\)，若是直接枚舉最壞是\(O(nkA)\)的可是跑不到，能過。ui
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int sg[N], n, k, a[N], vis[N];
int mex(int x) {
memset(vis, 0, sizeof(vis));
for(int i = 1; i <= n; ++i) {
if(x - a[i] >= 0) vis[sg[x - a[i]]] = 1;
}
for(int i = 0; i <= 100000; ++i) if(!vis[i]) return i;
}
int main() {
scanf("%d%d", &n, &k);
for(int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
}
for(int j = 1; j <= k; ++j) sg[j] = mex(j);
if(sg[k]) puts("First");
else puts("Second");
}
L
設\(f[l][r]\)表示按策略取完區間\([l,r]\)能夠獲得的分數。
直接按他們的策略模擬便可，這個策略是有階段性的。spa
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 3100;
ll f[N][N];
int n, a[N];
int main() {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
for(int i = 1; i <= n; ++i) if(n & 1) f[i][i] = a[i]; else f[i][i] = -a[i];
for(int len = 2; len <= n; ++len) {
for(int l = 1; l <= n; ++l) {
int r = l + len - 1;
if((n - len) & 1) f[l][r] = min(f[l + 1][r] - a[l], f[l][r - 1] - a[r]);
else f[l][r] = max(f[l + 1][r] + a[l], f[l][r - 1] + a[r]);
}
}
printf("%lld\n", f[1][n]);
}
M
用前綴和優化一下轉移便可。設計
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 100010;
const ll mod = 1e9 + 7;
int n, k, a[N];
ll f[101][N], sum[101][N];
/*
f[i][j]表示前i我的，一共分了j份。
*/
int main() {
int cur = 1;
scanf("%d%d", &n, &k);
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
sum[0][1] = f[0][1] = 1;
for(int i = 1; i <= k + 1; ++i) sum[0][i] = 1;
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= k + 1; ++j) {
(f[i][j] = sum[i - 1][j] - sum[i - 1][max(j - a[i] - 1, 0)]) %= mod;
sum[i][j] = (f[i][j] + sum[i][j - 1]) % mod;
}
}
printf("%lld\n", (f[n][k + 1] % mod + mod) % mod);
/* for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= k + 1; ++j) printf("%d ", sum[i][j]);
puts("");
}*/
}
N
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 500;
ll f[N][N], sum[N];
int n, a[N], ans = 0;
int main() {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]), sum[i] = sum[i - 1] + a[i];
memset(f, 0x3f, sizeof(f));
for(int i = 1; i <= n; ++i) f[i][i] = 0;
for(int len = 2; len <= n; ++len) {
for(int l = 1; l <= n; ++l) {
int r = l + len - 1;
if(r > n) break;
for(int k = l; k < r; ++k) {
f[l][r] = min(f[l][r], f[l][k] + f[k + 1][r] + sum[r] - sum[l - 1]);
}
}
}
printf("%lld\n", f[1][n]);
}
O
狀壓dp。
一開始寫了很傻的\(O(n^2 2^n)\)竟然都過了，\(O(n 2 ^n)\)的作法就是先枚舉集合，而後統計出當前匹配了多少人\(i\)，對\(f[i][S]\)枚舉轉移點轉移便可。code
#include <bits/stdc++.h>
using namespace std;
const int N = 22;
const int mod = 1e9 + 7;
int n, a[N][N];
int f[N][1 << N];
int main() {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= n; ++j) scanf("%d", &a[i][j]);
}
f[0][0] = 1;
for(int S = 1; S < (1 << n); ++S) {
int i = 0;
for(int j = 0; j < n; ++j) if((S >> j) & 1) ++i;
for(int j = 0; j < n; ++j) {
if(((S >> j) & 1) && a[i][j + 1]) (f[i][S] += f[i - 1][S ^ (1 << j)]) %= mod;
}
}
printf("%d\n", f[n][(1 << n) - 1]);
}
P
設\(f[i][0/1]\)表示節點\(i\)塗成黑色/白色的方案數。
有方程\(f[i][1]=\prod_{j\in son(i)} (f[j][0]+f[j][1]),f[i][0]=\sum_{j\in son(i)} f[j][1]\)排序
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
const int N = 100010;
int n;
ll f[N][2];
// f[i][0/1]表示將該節點塗成黑色/白色的方案數
int cnt, head[N];
struct edge { int to, nxt; } e[N << 1];
void ins(int u, int v) {
e[++cnt] = (edge) { v, head[u] };
head[u] = cnt;
}
void dfs(int u, int fa) {
f[u][0] = f[u][1] = 1;
for(int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if(v == fa) continue;
dfs(v, u);
(f[u][0] *= f[v][1]) %= mod;
(f[u][1] *= (f[v][0] + f[v][1]) % mod) %= mod;
}
}
int main() {
scanf("%d", &n);
for(int u, v, i = 1; i < n; ++i) {
scanf("%d%d", &u, &v);
ins(u, v), ins(v, u);
}
dfs(1, 0);
printf("%lld\n", (f[1][0] + f[1][1]) % mod);
}
Q
用BIT維護一下前綴\(\max\)轉移便可。it
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 200010;
int n, a[N], h[N];
ll c[N];
#define lowbit(i) (i & -i)
void add(int x, ll v) {
for(int i = x; i <= n; i += lowbit(i)) c[i] = max(c[i], v);
}
ll query(int x) {
ll ans = 0;
for(int i = x; i; i -= lowbit(i)) ans = max(ans, c[i]);
return ans;
}
int main() {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) scanf("%d", &h[i]);
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
for(int i = 1; i <= n; ++i) {
add(h[i], query(h[i]) + a[i]);
}
printf("%lld\n", query(n));
}
R
和bzoj cow relays同樣。
把矩陣乘法寫成相似floyd那樣，用矩陣快速冪維護便可。class
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 60;
const ll mod = 1e9 + 7;
int n;
ll K;
struct mat {
ll d[N][N];
mat() {memset(d, 0, sizeof(d));}
mat operator * (mat &x) {
mat ans;
for(int k = 1; k <= n; ++k) {
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= n; ++j) {
(ans.d[i][j] += d[i][k] * x.d[k][j] % mod) %= mod;
}
}
}
return ans;
}
} A[65];
int main() {
scanf("%d%lld", &n, &K);
for(int i = 1; i <= n; ++i) for(int j = 1; j <= n; ++j) scanf("%lld", &A[0].d[i][j]);
for(ll i = 1; (1LL << i) <= K; ++i) {
A[i] = A[i - 1] * A[i - 1];
}
mat ans;
for(int i = 1; i <= n; ++i) ans.d[i][i] = 1;
for(ll i = 0; (1LL << i) <= K; ++i)
if((K >> i) & 1LL) ans = ans * A[i];
ll sum = 0;
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= n; ++j) (sum += ans.d[i][j]) %= mod;
}
printf("%lld\n", sum);
}
S
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 100010;
const int mod = 1e9 + 7;
char s[N];
int d, n, a[N];
int f[10010][110];
int dfs(int cnt, int limit, int lead, int s) {
if(!cnt) return !s && !lead;
if(~f[cnt][s] && !limit && !lead) return f[cnt][s];
int sum = 0, ed = limit ? a[cnt] : 9;
for(int i = 0; i <= ed; ++i) {
(sum += dfs(cnt - 1, limit && (i == ed), lead && (!i), (s + i) % d)) %= mod;
}
if(!limit && !lead) return f[cnt][s] = sum;
return sum;
}
int main() {
memset(f, -1, sizeof(f));
scanf("%s%d", s + 1, &d);
n = strlen(s + 1);
for(int i = 1; i <= n; ++i) a[i] = s[i] - '0';
reverse(a + 1, a + n + 1);
printf("%lld\n", dfs(n, 1, 1, 0));
}
T
設\(f[i][j]\)表示填在第\(i\)位的數在已填入的數中排名爲\(j\)。
考慮這樣子設計狀態爲何是對的，若是第二維表示的是填入的數是\(j\)，那麼並不能知道前面\(i-1\)個數具體填的是什麼（由於排列不可重，因此不能填重複的數），也就無從轉移。
設成在已填入數中排名爲\(j\)，這樣知道的是相對關係，因此可轉移點也就肯定了（相對小於它的點）。
對於\(s[i]='<'\)，有\(f[i][j]=\sum_{k< j} f[i-1][k]\)
對於\(s[i]='>'\)，有\(f[i][j]=\sum_{j\le k< i}f[i-1][k]\)
維護一下前綴和便可\(O(n^2)\)轉移。
設計狀態的時候設計成相對關係有時候頗有用。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 3010;
const int mod = 1e9 + 7;
char s[N];
int f[N][N], sum[N], n;
/*
f[i][j]表示位置i，而後位置i的數在已填入的數中排名爲j
'>' f[i][j] = \sum_{k < j} f[i - 1][k]
'<' f[i][j] = \sum_{k >= j} f[i - 1][k]
*/
int main() {
scanf("%d", &n);
scanf("%s", s + 2);
f[1][1] = sum[1] = 1;
for(int i = 2; i <= n; ++i) {
for(int j = 1; j <= i; ++j) {
if(s[i] == '<') (f[i][j] += sum[j - 1]) %= mod;
else (f[i][j] += sum[i - 1] - sum[j - 1] + mod) %= mod;
}
for(int j = 1; j <= i; ++j) sum[j] = (sum[j - 1] + f[i][j]) % mod;
}
printf("%d\n", sum[n]);
}
U
狀壓dp。枚舉子集轉移便可。
\(f[S]=\sum f[S_1]+val[S\oplus S_1]\)其中\(S_1\)爲\(S\)的子集。
\(val[S]\)能夠\(O(n^22^n)\)處理出來。
複雜度爲\(O(3^n+n^22^n)\)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 17;
int n, a[N][N], b[N], cnt;
ll f[1 << N], val[1 << N];
int main() {
scanf("%d", &n);
for(int i = 0; i < n; ++i) {
for(int j = 0; j < n; ++j) scanf("%d", &a[i][j]);
}
for(int S = 1; S < (1 << n); ++S) {
cnt = 0;
for(int i = 0; i < n; ++i) {
if((S >> i) & 1) b[++cnt] = i;
}
for(int i = 1; i <= cnt; ++i) {
for(int j = i + 1; j <= cnt; ++j) val[S] += a[b[i]][b[j]];
}
}
for(int S = 1; S < (1 << n); ++S) {
for(int S1 = S; S1; S1 = (S1 - 1) & S) {
f[S] = max(f[S], f[S ^ S1] + val[S1]);
}
}
printf("%lld\n", f[(1 << n) - 1]);
}
V
W
線段樹優化dp。
由於若是選了一個點，那麼右端點在它以前的線段顯然不會對這個點有影響，那麼把那些線段的貢獻都加上去以後取max轉移便可。具體可見代碼。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 300010;
const ll inf = 1e16;
ll f[N];
int n, m;
struct tree { int l, r; ll mx, tag; } t[N << 2];
struct line { int l, r; ll v; } a[N];
#define lc (rt << 1)
#define rc (rt << 1 | 1)
void build(int l, int r, int rt) {
t[rt].l = l; t[rt].r = r;
if(l == r) return; int mid = (l + r) >> 1;
build(l, mid, lc); build(mid + 1, r, rc);
}
void up(int rt) { t[rt].mx = max(t[lc].mx, t[rc].mx); }
#define l (t[rt].l)
#define r (t[rt].r)
#define mid ((l + r) >> 1)
void down(int rt) {
if(t[rt].tag) {
t[lc].mx += t[rt].tag; t[rc].mx += t[rt].tag;
t[lc].tag += t[rt].tag; t[rc].tag += t[rt].tag;
t[rt].tag = 0;
}
}
void upd(int L, int R, ll c, int rt) {
if(L <= l && r <= R) {
t[rt].mx += c; t[rt].tag += c;
return;
}
down(rt);
if(L <= mid) upd(L, R, c, lc);
if(R > mid) upd(L, R, c, rc);
up(rt);
}
ll query(int L, int R, int rt) {
if(L <= l && r <= R) return t[rt].mx;
down(rt); ll ans = -inf;
if(L <= mid) ans = max(ans, query(L, R, lc));
if(R > mid) ans = max(ans, query(L, R, rc));
return ans;
}
#undef l
#undef r
#undef mid
#undef lc
#undef rc
bool operator < (line a, line b) { return a.r < b.r; }
int main() {
scanf("%d%d", &n, &m);
build(1, n, 1);
for(int i = 1; i <= m; ++i) {
scanf("%d%d%lld", &a[i].l, &a[i].r, &a[i].v);
}
sort(a + 1, a + m + 1);
int r = 1;
for(int i = 1; i <= n; ++i) {
upd(i, i, query(1, i, 1), 1);
while(a[r].r == i) {
upd(a[r].l, a[r].r, a[r].v, 1);
++r;
}
}
printf("%lld\n", max(0LL, query(1, n, 1)));
}
X
按\(w+s\)升序排序而後相似揹包轉移便可。
這個排序的貪心能夠用排序不等式證實。（也能夠感性理解）
#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
typedef long long ll;
int n;
ll f[N][N * 30];
struct Node { int w, s; ll v; } a[N];
/*
設f[i][j]表示前i個，而後總重量爲j的最大答案。
*/
bool operator < (Node a, Node b) {
if(a.s + a.w == b.s + b.w) return a.s < b.s;
return a.s + a.w < b.s + b.w;
}
int main() {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) {
scanf("%d%d%lld", &a[i].w, &a[i].s, &a[i].v);
}
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; ++i) {
for(int j = 0; j <= a[n].s + a[n].w; ++j) f[i][j] = max(f[i][j], f[i - 1][j]);
for(int j = 0; j <= a[i].s; ++j) f[i][j + a[i].w] = max(f[i][j + a[i].w], f[i - 1][j] + a[i].v);
}
ll ans = 0;
for(int i = 0; i <= a[n].s + a[n].w; ++i) ans = max(ans, f[n][i]);
printf("%lld\n", ans);
}
Y
計數dp。首先一個沒有任何障礙的\(h\times w\)的網格圖從\((1,1)\)走到\((h,w)\)的方案數爲\(C^{h-1}_{h+w-2}\)（考慮一共要走\(h-1\)次向下，\(w-1\)次向右，一共\(h+w-2\)次操做）
設\(f[i]\)表示僅通過第\(i\)個黑點且目前位於第\(i\)個黑點的方案數。
將黑點按\(x\)爲第一關鍵字，\(y\)爲第二關鍵字升序排序。
有\(f[i]=C^{x_i-1}_{x_i+y_i-2}-\sum_{j<i,x_i\le x_j,y_i\le y_j}f[j]\times C^{x_i-x_j}_{x_i-x_j+y_i-y_j}\)
欽定\((h,w)\)爲第\(n+1\)個黑點，那麼答案就是\(f[n+1]\)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
const int N = 200010;
ll fac[N], inv[N], f[N];
int h, w, n;
struct Node { int x, y; } a[N];
ll power(ll a, ll b) {
ll ans = 1;
while(b) {
if(b & 1) ans = ans * a % mod;
a = a * a % mod; b >>= 1;
} return ans;
}
bool operator < (Node a, Node b) {
if(a.x == b.x) return a.y < b.y;
return a.x < b.x;
}
ll C(int n, int m) {
return fac[m] * inv[n] % mod * inv[m - n] % mod;
}
int main() {
scanf("%d%d%d", &h, &w, &n);
fac[0] = inv[0] = 1;
for(int i = 1; i < N; ++i) {
fac[i] = 1LL * fac[i - 1] * i % mod;
inv[i] = power(fac[i], mod - 2);
}
for(int i = 1; i <= n; ++i) {
scanf("%d%d", &a[i].x, &a[i].y);
}
sort(a + 1, a + n + 1);
a[n + 1] = (Node) { h, w };
for(int i = 1; i <= n + 1; ++i) {
f[i] = C(a[i].x - 1, a[i].x + a[i].y - 2);
for(int j = 1; j < i; ++j) {
if(a[j].x <= a[i].x && a[j].y <= a[i].y) {
f[i] -= C(a[i].x - a[j].x, a[i].x - a[j].x + a[i].y - a[j].y) * f[j] % mod;
(f[i] += mod) %= mod;
}
}
}
printf("%lld\n", f[n + 1]);
}
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