CF 96 D. Volleyball

D. Volleyball

http://codeforces.com/contest/96/problem/D

题意：

　　n个路口，m条双向路，每条长度为w。每个路口有一个出租车司机，最多可以乘坐这辆车走长度只要坐他的车，就必须交c元，最多可以载你走的长度为t的路。问从x到y的最小花费是多少。

分析：

　　第一遍SPFA求出每个点它能到的所有点，边权就是乘坐出租车的费用，第二遍直接跑最短路。稀疏图用了spfa

代码：

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<iostream>
 #include<cctype>
 #include<set>
 #include<vector>
 #include<queue>
 #include<map>
 #define fi(s) freopen(s,"r",stdin);
 #define fo(s) freopen(s,"w",stdout);
 using namespace std;
 typedef long long LL;
 
 inline int read() {
     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }
 
 const int N = ;
 
 struct ShortestPath{
     int head[N], len[N * N], nxt[N * N], to[N * N], q[N * N], En;
     LL dis[N], INF;
     bool vis[N];
     void add_edge(int u,int v,int w) {
         ++En; to[En] = v; len[En] = w; nxt[En] = head[u]; head[u] = En;
     }
     void spfa(int S) {
         memset(dis, 0x3f, sizeof(dis)); INF = dis[];
         int L = , R = ;
         dis[S] = ; q[++R] = S; vis[S] = true;
         while (L <= R) {
             int u = q[L ++]; vis[u] = false;
             for (int i = head[u]; i; i = nxt[i]) {
                 int v = to[i];
                 if (dis[v] > dis[u] + len[i]) {
                     dis[v] = dis[u] + len[i];
                     if (!vis[v]) q[++R] = v, vis[v] = true;
                 }
             }
         }
     }
 }G1, G2;
 
 int main() {
     int n = read(), m = read(), S = read(), T = read();
     for (int i = ; i <= m; ++i) {
         int u = read(), v = read(), w = read();
         G1.add_edge(u, v, w); G1.add_edge(v, u, w);
     }
     for (int i = ; i <= n; ++i) {
         int d = read(), c = read();
         G1.spfa(i);
         for (int j = ; j <= n; ++j) {
             if (G1.dis[j] <= d && i != j) G2.add_edge(i, j, c); // 此处为单向边
         }
     }
     G2.spfa(S);
     if (G2.dis[T] == G2.INF) puts("-1");
     else cout << G2.dis[T];
     return ;
 }