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=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 1005;

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[0]; 
        int L = 1, R = 0; 
        dis[S] = 0; 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 = 1; 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 = 1; i <= n; ++i) {
        int d = read(), c = read();
        G1.spfa(i);
        for (int j = 1; 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 0;
}