Codeforces Round #302 (Div. 1) B

B - Destroying Roads

思路：这么菜的题我居然想了40分钟。。。 n^2枚举两个交汇点，点与点之间肯定都跑最短路，取最小值。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define y1 skldjfskldjg
#define y2 skldfjsklejg

using namespace std;

const int N = 3000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;

int d[N][N], n, m, s1, t1, l1, s2, t2, l2;
vector<int> edge[N];
void Dij(int S, int d[N]) {
    queue<int> que;
    d[S] = 0; que.push(S);
    while(!que.empty()) {
       int u = que.front(); que.pop();
       for(int v : edge[u]) {
            if(~d[v]) continue;
            d[v] = d[u] + 1;
            que.push(v);
       }
    }
}

int cal(int u, int v) {
    int ret1 = min(d[s1][u] + d[t1][v], d[t1][u] + d[s1][v]);
    int ret2 = min(d[s2][u] + d[t2][v], d[t2][u] + d[s2][v]);
    if(ret1 + d[u][v] > l1 || ret2 + d[u][v] > l2) return inf;

    return d[u][v] + ret1 + ret2;
}

int main() {
    memset(d, -1, sizeof(d));
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i++) {
        int u, v;
        scanf("%d%d", &u, &v);
        edge[u].push_back(v);
        edge[v].push_back(u);
    }
    for(int i = 1; i <= n; i++) Dij(i, d[i]);
    scanf("%d%d%d%d%d%d", &s1, &t1, &l1, &s2, &t2, &l2);

    if(d[s1][t1] > l1 || d[s2][t2] > l2) {
        puts("-1");
        return 0;
    }
    int ans = d[s1][t1] + d[s2][t2];

    for(int u = 1; u <= n; u++) {
        for(int v = 1; v <= n; v++) {
            ans = min(ans, cal(u, v));
        }
    }
    printf("%d
", m - ans);
    return 0;
}

/*
*/