Codeforces Round #302 (Div. 2) D

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题目链接:

http://codeforces.com/contest/544/problem/D

题意:

有n个城镇，m条边权为1的双向边
让你破坏最多的道路，使得从s1到t1，从s2到t2的距离分别不超过l1和l2

题解:

跑一发最短路，然后最后留下的图肯定是出了s1-t1,s2-t2这两条路之外，其他路都被删除了
那么我们枚举重叠的道路就好了【可能反向】

代码:

#include <bits/stdc++.h>using namespace std;typedef long long ll;#define MS(a) memset(a,0,sizeof(a))#define MP make_pair#define PB push_backconst int INF = 0x3f3f3f3f;const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;inline ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}//////////////////////////////////////////////////////////////////////////const int maxn = 3e3+10;int inq[maxn],d[maxn][maxn];vector<int> E[maxn];int main(){    int n=read(),m=read();    for(int i=0; i<m; i++){        int u=read(),v=read();        E[u].push_back(v);        E[v].push_back(u);    }    int s1,t1,l1,s2,t2,l2;    scanf("%d%d%d%d%d%d",&s1,&t1,&l1,&s2,&t2,&l2);    for(int i=1; i<=n; i++){        MS(inq);        for(int j=0; j<=n; j++) d[i][j] = INF;        queue<int> q;        q.push(i),inq[i]=1,d[i][i]=0;        while(!q.empty()){            int now = q.front();            q.pop(),inq[now] = 0;            for(int k=0; k<(int)E[now].size(); k++){                int v = E[now][k];                if(d[i][v] > d[i][now]+1){                    d[i][v] = d[i][now]+1;                    if(inq[v]) continue;                    inq[v] = 1;                    q.push(v);                }            }        }    }    if(d[s1][t1]>l1 || d[s2][t2]>l2){        cout << -1 << endl;        return 0;    }    int ans = d[s1][t1]+d[s2][t2];    for(int i=1; i<=n; i++)        for(int j=1; j<=n; j++){            if(d[s1][i]+d[i][j]+d[j][t1]<=l1 && d[s2][i]+d[i][j]+d[j][t2]<=l2)                ans = min(ans,d[s1][i]+d[i][j]+d[j][t1]+d[s2][i]+d[j][t2]);            if(d[s1][i]+d[i][j]+d[j][t1]<=l1 && d[t2][i]+d[i][j]+d[j][s2]<=l2) // 反向的时候                ans = min(ans,d[s1][i]+d[i][j]+d[j][t1]+d[t2][i]+d[j][s2]);        }    cout << m-ans << endl;    return 0;}/*10 111 32 33 44 54 63 73 84 94 107 98 101 5 36 2 3*/

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