直接BFSnode
#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ') #define enter putchar('\n') #define eps 1e-10 #define MAXN 100005 //#define ivorysi using namespace std; typedef long long int64; typedef unsigned int u32; typedef double db; template<class T> void read(T &res) { res = 0;T f = 1;char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { res = res * 10 +c - '0'; c = getchar(); } res *= f; } template<class T> void out(T x) { if(x < 0) {x = -x;putchar('-');} if(x >= 10) { out(x / 10); } putchar('0' + x % 10); } int N,M; char s[1005][1005]; int dis[1005][1005]; bool vis[1005][1005]; queue<pii > Q; int dx[] = {0,-1,0,1}; int dy[] = {1,0,-1,0}; void Solve() { read(N);read(M); for(int i = 1 ; i <= N ; ++i) { scanf("%s",s[i] + 1); for(int j = 1 ; j <= M ; ++j) { if(s[i][j] == '#') { Q.push(mp(i,j)); vis[i][j] = 1; } } } int ans = 0; while(!Q.empty()) { pii u = Q.front();Q.pop(); ans = max(dis[u.fi][u.se],ans); for(int k = 0 ; k < 4 ; ++k) { int tx = u.fi + dx[k],ty = u.se + dy[k]; if(tx >= 1 && tx <= N && ty >= 1 && ty <= M) { if(!vis[tx][ty]) { vis[tx][ty] = 1; dis[tx][ty] = dis[u.fi][u.se] + 1; Q.push(mp(tx,ty)); } } } } out(ans);enter; } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); }
我15min切了第一題c++
第二題想了兩個點函數
首先是分紅兩維ui
以\(L,R\)爲例,若是我想從最右邊出去,我確定本身不能用Lspa
可是L會對後手的操做有限制,處理出一個後綴表示後綴中先手使用L後手補救L,往左最多能移動幾個code
當我想拼命使用R時,後手若是左移必須在後綴中先手使用L的峯值以前ci
而後分紅四種,寫成一個函數就好………………rem
掉分了,難過get
#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ') #define enter putchar('\n') #define eps 1e-10 #define MAXN 200005 //#define ivorysi using namespace std; typedef long long int64; typedef unsigned int u32; typedef double db; template<class T> void read(T &res) { res = 0;T f = 1;char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { res = res * 10 +c - '0'; c = getchar(); } res *= f; } template<class T> void out(T x) { if(x < 0) {x = -x;putchar('-');} if(x >= 10) { out(x / 10); } putchar('0' + x % 10); } int H,W,N; int sr,sc; char s[MAXN],t[MAXN]; int suf[MAXN]; bool Process(char a,char b,int dis,int lim) { suf[N + 1] = 0;int sum = 0; for(int i = N ; i >= 1 ; --i) { suf[i] = suf[i + 1]; if(s[i] == b) { suf[i] = min(suf[i],sum - 1); --sum; } if(t[i] == a) { if(sum < 0) ++sum; } } int pos = dis; for(int i = 1 ; i <= N ; ++i) { if(s[i] == a) ++pos; if(pos > lim) return false; if(t[i] == b) { if(pos - 1 + suf[i + 1] >= 1) --pos; } } return true; } void Solve() { read(H);read(W);read(N); read(sr);read(sc); scanf("%s",s + 1);scanf("%s",t + 1); if(!Process('D','U',sr,H) || !Process('U','D',H + 1 - sr,H) || !Process('R','L',sc,W) || !Process('L','R',W + 1 - sc,W)) { puts("NO"); } else puts("YES"); } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); }
又是考後兩分鐘想出題解,哭了it
由於B題沒切C題看的心煩意亂
每次至關於割掉全部葉子,若是選的根是一個葉子則保留一個葉子
若是是一條鏈,長度爲1後手勝,長度爲2先手勝,長度爲3先手勝,長度爲4後手勝,是這樣一個循環……
對應到樹上呢,很神奇,是樹的直徑!
由於咱們對於直徑也有一次割兩個或者一次割一個的操做……
就……作完了
#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ') #define enter putchar('\n') #define eps 1e-10 #define MAXN 200005 //#define ivorysi using namespace std; typedef long long int64; typedef unsigned int u32; typedef double db; template<class T> void read(T &res) { res = 0;T f = 1;char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { res = res * 10 +c - '0'; c = getchar(); } res *= f; } template<class T> void out(T x) { if(x < 0) {x = -x;putchar('-');} if(x >= 10) { out(x / 10); } putchar('0' + x % 10); } struct node { int to,next; }E[MAXN * 2]; int sumE,head[MAXN],N; int dis[MAXN]; void add(int u,int v) { E[++sumE].to = v; E[sumE].next = head[u]; head[u] = sumE; } void dfs(int u,int fa) { dis[u] = dis[fa] + 1; for(int i = head[u] ; i ; i = E[i].next) { int v = E[i].to; if(v != fa) { dfs(v,u); } } } void Solve() { read(N); if(N == 1) { puts("First");return; } int a,b; for(int i = 1 ; i < N ; ++i) { read(a);read(b); add(a,b);add(b,a); } dfs(1,0); int u = 1; for(int i = 1 ; i <= N ; ++i) { if(dis[i] > dis[u]) u = i; } dfs(u,0); u = 1; for(int i = 1 ; i <= N ; ++i) { if(dis[i] > dis[u]) u = i; } if(dis[u] % 3 == 2) {puts("Second");} else puts("First"); } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); }
答案確定不會超過\(log(HW)\),最可能是15
記錄計算\((r,c)\)到\((r' ,c)\)這兩個點最右能延伸到哪一列
以及\((r,c)\)到\((r,c')\)最下能延伸到哪一行
橫豎能夠交替轉移
狀態壓縮一下空間就是\(n^3\)
複雜度的話多是\(n^4\)可是有個小常數
#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ') #define enter putchar('\n') #define eps 1e-10 #define MAXN 205 //#define ivorysi using namespace std; typedef long long int64; typedef unsigned int u32; typedef double db; template<class T> void read(T &res) { res = 0;T f = 1;char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { res = res * 10 +c - '0'; c = getchar(); } res *= f; } template<class T> void out(T x) { if(x < 0) {x = -x;putchar('-');} if(x >= 10) { out(x / 10); } putchar('0' + x % 10); } int H,W,sum[MAXN][MAXN]; int f[2][MAXN][MAXN][MAXN],g[2][MAXN][MAXN][MAXN]; char s[MAXN][MAXN]; bool check(int xa,int ya,int xb,int yb) { if(yb > W || xb > H) return false; int all = (yb - ya + 1) * (xb - xa + 1); int t = sum[xb][yb] + sum[xa - 1][ya - 1] - sum[xb][ya - 1] - sum[xa - 1][yb]; if(t == all || t == 0) return true; return false; } void Solve() { read(H);read(W); for(int i = 1 ; i <= H ; ++i) { scanf("%s",s[i] + 1); for(int j = 1 ; j <= W ; ++j) sum[i][j] = (s[i][j] == '#'); for(int j = 1 ; j <= W ; ++j) sum[i][j] += sum[i][j - 1]; } for(int i = 1 ; i <= H ; ++i) { for(int j = 1 ; j <= W ; ++j) sum[i][j] += sum[i - 1][j]; } int cur = 0; for(int i = 1 ; i <= H ; ++i) { for(int j = i ; j <= H ; ++j) { for(int c = 1 ; c <= W ; ++c) { int t = f[cur][i][j][c - 1]; if(check(i,c,j,c)) t = max(t,c); if(t >= c) { while(check(i,c,j,t + 1)) ++t; } else t = c - 1; f[cur][i][j][c] = t; } } } for(int i = 1 ; i <= W ; ++i) { for(int j = i ; j <= W ; ++j) { for(int r = 1 ; r <= H ; ++r) { int t = g[cur][i][j][r - 1]; if(check(r,i,r,j)) t = max(t,r); if(t >= r) { while(check(r,i,t + 1,j)) ++t; } else t = r - 1; g[cur][i][j][r] = t; } } } for(int k = 0 ; k <= 20 ; ++k) { if(f[cur][1][H][1] >= W || g[cur][1][W][1] >= H) { out(k);enter;return; } memset(f[cur ^ 1],0,sizeof(f[cur ^ 1])); memset(g[cur ^ 1],0,sizeof(g[cur ^ 1])); for(int i = 1 ; i <= H ; ++i) { for(int j = i ; j <= H ; ++j) { for(int c = 1 ; c <= W ; ++c) { int t = f[cur][i][j][c]; if(t >= W) {f[cur ^ 1][i][j][c] = W;continue;} t = f[cur][i][j][t + 1]; while(1) { int k = g[cur][c][t + 1][i]; k = g[cur][c][t + 1][k + 1]; if(k >= j) ++t; else break; } f[cur ^ 1][i][j][c] = t; } } } for(int i = 1 ; i <= W ; ++i) { for(int j = i ; j <= W ; ++j) { for(int r = 1 ; r <= H ; ++r) { int t = g[cur][i][j][r]; if(t >= H) {g[cur ^ 1][i][j][r] = H;continue;} t = g[cur][i][j][t + 1]; while(1) { int k = f[cur][r][t + 1][i]; k = f[cur][r][t + 1][k + 1]; if(k >= j) ++t; else break; } g[cur ^ 1][i][j][r] = t; } } } cur ^= 1; } } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); return 0; }
……E題咋也那麼簡單
失去漲分良機
這個就是,不妨設第一次出現的字母是R,那麼B確定不會在環上連續出現兩次
咱們斷開B,就造成了一個個小島
小島的長度必須是奇數,由於若是是偶數則中心點走到兩邊只有奇數,而端點到端點就只有偶數,而第一段連續R不可能又是奇數又是偶數
若是第一段連續的R長度是t,那麼小島的長度最大是t+1,不然小島的長度最大是t
而中間的B呢,至關於咱們更換小島,或者呆在原來的小島上,可是因爲每一個點都會經歷中間的B,因此就算換了小島,每一個小島的兩邊都會有點
咱們要走偶數個R,只須要來回走一條邊
而奇數個點,咱們要走到小島的另外一邊,這個就對小島長度有了限制
注意最後一段R不限制咱們的小島
這樣的話就能夠dp了
特殊狀況是若是全是R那麼小島的長度能夠無窮大
(我爲啥要起小島這樣的名字呢2333)
#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ') #define enter putchar('\n') #define eps 1e-10 #define MAXN 200005 //#define ivorysi using namespace std; typedef long long int64; typedef unsigned int u32; typedef double db; template<class T> void read(T &res) { res = 0;T f = 1;char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { res = res * 10 +c - '0'; c = getchar(); } res *= f; } template<class T> void out(T x) { if(x < 0) {x = -x;putchar('-');} if(x >= 10) { out(x / 10); } putchar('0' + x % 10); } const int MOD = 1000000007; int N,M,L; char s[MAXN]; bool flag = 0; int f[MAXN][2],dp[MAXN],sum[MAXN][2]; int inc(int a,int b) { return a + b >= MOD ? a + b - MOD : a + b; } int mul(int a,int b) { return 1LL * a * b % MOD; } void update(int &x,int y) { x = inc(x,y); } void Solve1() { int ans = 0; f[1][1] = 1;f[1][0] = 1; for(int i = tijie2 ; i <= N ; ++i) { f[i][0] = f[i - 1][1]; f[i][1] = inc(f[i - 1][1],f[i - 1][0]); } ans = inc(f[N][0],f[N][1]); memset(f,0,sizeof(f)); f[1][0] = 1; for(int i = 2 ; i < N ; ++i) { f[i][0] = f[i - 1][1]; f[i][1] = inc(f[i - 1][1],f[i - 1][0]); } ans = inc(ans,MOD - f[N - 1][1]); out(ans);enter; } void Solve2() { dp[0] = 1;sum[0][0] = 1; for(int i = 1 ; i <= N ; ++i) { if(i >= 2) { int t = sum[i - 2][i & 1]; if(i > L + 1) update(t,MOD - sum[i - L - 2][i & 1]); dp[i] = t; } sum[i][0] = sum[i - 1][0];sum[i][1] = sum[i - 1][1]; sum[i][i & 1] = inc(sum[i][i & 1],dp[i]); } int ans = dp[N]; for(int i = 1 ; i <= L ; i += 2) { update(ans,mul(i,dp[N - i - 1])); } out(ans);enter; } void Init() { read(N);read(M); scanf("%s",s + 1); while(M >= 1 && s[M] == s[1]) --M; L = N; if(M == 0) flag = 1; else { for(int i = 1 ; i <= M ; ++i) { if(s[i] != s[1]) continue; else { int j = i; while(s[j + 1] == s[1]) ++j; int t = j - i + 1; if(i == 1) { if(t % 2 == 0) L = min(L,t + 1); else L = min(L,t); } else { if(t % 2 == 1) L = min(L,t); } i = j; } } } if(L == N) flag = 1; if(flag) Solve1(); else Solve2(); } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Init(); }
彷佛就是個暴力?
咱們只要每次把存在一條邊\((a,b)\),c在\((a,b)\)的路徑上,且存在\((a,c)\),刪掉\((a,b)\),加上\((b,c)\),最後只須要對於每一個點當作根,在這些路徑上走從祖先到兒子的邊,不然不走,每一個點搜出來的點的個數(除了本身)就是這個點能夠連出去的邊的總數
怎麼搜,就是直接加一條邊的時候,找出這條鏈,而後找這條鏈兩端看能不能縮小一個範圍,而後新加的邊和兩個端點以前的邊比較一下,能刪則刪
#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ') #define enter putchar('\n') #define eps 1e-10 #define MAXN 2005 //#define ivorysi using namespace std; typedef long long int64; typedef unsigned int u32; typedef double db; template<class T> void read(T &res) { res = 0;T f = 1;char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { res = res * 10 +c - '0'; c = getchar(); } res *= f; } template<class T> void out(T x) { if(x < 0) {x = -x;putchar('-');} if(x >= 10) { out(x / 10); } putchar('0' + x % 10); } struct node { int to,next; }E[MAXN * 2]; int N,M,sumE,head[MAXN],dis[MAXN][MAXN]; int fa[MAXN][MAXN]; bool vis[MAXN][MAXN]; bool g[MAXN][MAXN]; set<int> S[MAXN]; void add(int u,int v) { E[++sumE].to = v; E[sumE].next = head[u]; head[u] = sumE; } void dfs(int id,int u) { for(int i = head[u] ; i ; i = E[i].next) { int v = E[i].to; if(v != fa[id][u]) { fa[id][v] = u; dis[id][v] = dis[id][u] + 1; dfs(id,v); } } } vector<int> getline(int a,int b) { vector<int> res; while(a != b) { res.pb(a); a = fa[b][a]; } res.pb(b); return res; } void build(int a,int b) { vis[a][b] = vis[b][a] = 1; S[a].insert(b);S[b].insert(a); } void destroy(int a,int b) { vis[a][b] = vis[b][a] = 0; S[a].erase(b);S[b].erase(a); } void AddE(int a,int b) { if(vis[a][b]) return; vector<int> v = getline(a,b); int s = v.size() - 1; int f = a,t = b,q = 0; for(int i = 0 ; i < s ; ++i) { if(vis[f][v[i]]) {f = v[i];q = i;} } for(int i = s - 1 ; i >= q ; --i) { if(vis[t][v[i]]) t = v[i]; } if(t == a) return; v.clear(); if(vis[f][t]) return; vector<pii> E; for(auto k : S[f]) { if(dis[f][k] == dis[f][t] + dis[t][k]) { E.pb(mp(t,k)); v.pb(k); } } for(auto k : v) destroy(k,f); v.clear(); for(auto k : S[t]) { if(dis[t][k] == dis[t][f] + dis[f][k]) { E.pb(mp(f,k)); v.pb(k); } } for(auto k : v) destroy(k,t); v.clear(); build(f,t); for(auto k : E) { AddE(k.fi,k.se); } } void find(int id,int u) { g[id][u] = 1; for(auto t : S[u]) { if(g[id][t]) continue; if(dis[t][id] == dis[id][u] + dis[u][t]) find(id,t); } } void Solve() { read(N);read(M); int a,b; for(int i = 1 ; i < N ; ++i) { read(a);read(b); add(a,b);add(b,a); } for(int i = 1 ; i <= N ; ++i) dfs(i,i); for(int i = 1 ; i <= M ; ++i) { read(a);read(b); AddE(a,b); } int ans = 0; for(int i = 1 ; i <= N ; ++i) { find(i,i); for(int j = i + 1 ; j <= N ; ++j) if(g[i][j]) ++ans; } out(ans);enter; } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); return 0; }
明天期中了,rp++
(但願別墊底,枯了)
ps:我發現yutaka的題故意會把B題放的難一點。。。