[PA2012] Tax

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題意：給出一個N個點M條邊的無向圖，通過一個點的代價是進入和離開這個點的兩條邊的邊權的較大值，求從起點1到點N的最小代價。起點的代價是離開起點的邊的邊權，終點的代價是進入終點的邊的邊權N<=100000 M<=200000

解題思路

難免要吐槽一下這題的數據，久調一下午無果與標程對拍沒有任何差錯不知道爲何就是WA 既然極限數據已經和標程拍上了那麼權當出了吧……

不過仍是一道好題

首先考慮這道題暴力的作法——將每條邊做爲新圖的點，而後枚舉原圖的點，遍歷一遍這個點相鄰的全部邊，按照大小打擂在新圖中連邊，可是這樣邊的數量多達$M^2$

能夠考慮優化邊的數量，用到差分的思想——以其中一條邊做爲基準，往上走要加，往下走不加。做爲基準的這一條邊也就是當前路徑的入邊，至於出邊，只須要沿着差分的邊走就能夠了。因而咱們所須要作的就是將每一個點相鄰的全部邊排序，而且相鄰的連邊——大的往小的權值爲0，小的往大的權值爲差值。而且對於每一條邊，它的反向邊應當與它連一條權值爲其自己的邊，做爲基準嘛

Code

/*By DennyQi 2018.8.11*/ #include <cstdio> #include <queue> #include <cstring> #include <algorithm>
#define  r  read()
#define  lr lread()
#define  Max(a,b)  (((a)>(b)) ? (a) : (b))
#define  Min(a,b)  (((a)<(b)) ? (a) : (b))
using namespace std; typedef long long ll; const int MAXN = 2000010; const int MAXM = 2000010; const int INF = 1e18; inline int read(){ int x = 0; int w = 1; register int c = getchar(); while(c ^ '-' && (c < '0' || c > '9')) c = getchar(); if(c == '-') w = -1, c = getchar(); while(c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar(); return x * w; } inline int lread(){ ll x = 0; int w = 1; register int c = getchar(); while(c ^ '-' && (c < '0' || c > '9')) c = getchar(); if(c == '-') w = -1, c = getchar(); while(c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar(); return x * w; } struct Edge{ ll len; int idx; }e[MAXM]; struct Dij{ ll w; int idx; }; inline bool operator < (const Dij& a, const Dij& b){ return a.w > b.w; } int N,M,x,y,S,T,v,top; int pfirst[MAXM],pnxt[MAXM],pto[MAXM],pnum_edge=-1; ll pcost[MAXM],cost[MAXM],z; int first[MAXM],nxt[MAXM],to[MAXM],num_edge=-1; ll d[MAXM]; bool vis[MAXM]; priority_queue <Dij> q; inline bool comp(const Edge& a, const Edge& b){ return a.len < b.len; } inline void add(int u, int v, int w){ // printf("%lld->%lld(%lld)\n",u,v,w);
    to[++num_edge] = v; cost[num_edge] = w; nxt[num_edge] = first[u]; first[u] = num_edge; } inline void padd(int u, int v, int w){ pto[++pnum_edge] = v; // printf("num(%lld): %lld->%lld(%lld)\n",pnum_edge,u,v,w);
    pcost[pnum_edge] = w; pnxt[pnum_edge] = pfirst[u]; pfirst[u] = pnum_edge; if(u == 1){ add(S, pnum_edge, w); } if(v == N){ add(pnum_edge, T, w); } } inline void Dijkstra(int s){ for(int i = 0; i <= T; ++i) d[i] = INF; d[s] = 0; q.push((Dij){0,s}); ll u,v; while(!q.empty()){ u = q.top().idx; q.pop(); if(vis[u]) continue; vis[u] = 1; for(int i = first[u]; i != -1; i = nxt[i]){ v = to[i]; if(d[u] + cost[i] < d[v]){ d[v] = d[u] + cost[i]; q.push((Dij){d[v],v}); } } } } int main(){ // freopen(".in","r",stdin); // freopen("qxz.out","w",stdout);
    N = r, M = r; memset(pfirst,-1,sizeof(pfirst)); memset(first,-1,sizeof(first)); S = M*2+1; T = M*2+2; // printf("S = %lld T = %lld\n",S,T);
    for(int i = 1; i <= M; ++i){ x = r, y = r, z = lr; padd(x, y, z); padd(y, x, z); } int v; for(int x = 2; x < N; ++x){ top = 0; for(int i = pfirst[x]; i != -1; i = pnxt[i]){ e[++top] = (Edge){pcost[i], i}; } sort(e+1,e+top+1,comp); for(int i = 1; i <= top; ++i){ if(i < top){ add(e[i].idx, e[i+1].idx, e[i+1].len-e[i].len); } if(i > 1){ add(e[i].idx, e[i-1].idx, 0); } add(e[i].idx^1, e[i].idx, e[i].len); } } Dijkstra(S); printf("%lld", d[T]); return 0; }