Codeforces 544D Destroying Roads

链接：http://codeforces.com/problemset/problem/543/B

                    B. Destroying Roads

time limit per test:2 seconds

In some country there are exactly n cities and m bidirectional roads

connecting the cities.Cities are numbered with integers from 1 to n.

If cities a and b are connected by a road,then in an hour you can go

along this road either from city a to city b, or from city b to city a.

The road network is such that from any city you can get to any other

one by moving alongthe roads.

You want to destroy the largest possible number of roads in the country

so that the remainingroads would allow you to get from city s1 to city t1 

in at most l1 hours and get from city s2 tocity t2 in at most l2 hours.

Determine what maximum number of roads you need to destroy in order

to meet the conditionof your plan. If it is impossible to reach the desired

result, print -1.

Input

The first line contains two integers n, m (1 ≤ n ≤ 3000

, ) —the number of cities and roads

in the country, respectively.Next m lines contain the descriptions of the

roads as pairs of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi).

It is guaranteed that the roads that are given in the description can

transport you from any cityto any other one. It is guaranteed that each

pair of cities has at most one road between them.

The last two lines contains three integers each, s1, t1, l1 

and s2, t2, l2, respectively (1 ≤ si, ti ≤ n, 0 ≤ li ≤ n).

Output

    Print a single number — the answer to the problem.

If the it is impossible to meet the conditions,print -1

题意：

        给出一张n个点，m条边的无向图，问最多能删除几条边，

同时保证给定点之间的距离<=某一值。

       or 给定n个点m条边的无向图(边权全为1)，让你去掉最多的边使得d(s1, t1) <= l1

&& d(s2, t2) <= l2，若不能满足输出-1，反之输出可以去掉的最多边数。

Examples

思路：

        例如给出的两条限制路径为a->...->b，c->...->d。如果这两条路径不相交的话，

显然去掉的最多边数=m-dist【a】【b】-dist【c】【d】。dist可以用spfa求一下最短路，

边权均为1。别的肯定会都超时，spfa有不超时的可能。

        然后当两条路径之间有公共路径时，我们将公共路径作为中间点更新最小保留边数，

又因为n只有10^3，所以可以两重循环枚举所有边，不是最短路中公共边的边产生的

答案肯定要比最优解大的，所以只是多几个无用的冗余判断，不影响最终结果。其次

注意的是四个点之间的不同的位置关系有两种，分别更新一下答案即可。

input
5 4
1 2
2 3
3 4
4 5
1 3 2
3 5 2
output
0
input
5 4
1 2
2 3
3 4
4 5
1 3 2
2 4 2
output
1
input
5 4
1 2
2 3
3 4
4 5
1 3 2
3 5 1
output
-1

 

AC代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define N 3005
#define inf 0x3f3f3f3f
struct Edge
{
    int to,next,w;
}edge[N*N];
int cnt,n,head[N],low[N][N],w[N];
bool vis[N];
inline void add(int u,int v,int w)
{
    edge[cnt].to=v;
    edge[cnt].w=w;
    edge[cnt].next=head[u];
    head[u]=cnt++;
}
void spfa()
{
    int i,j;
    for(j=1; j<=n; ++j){
        memset(vis,0,sizeof(vis));
        low[j][j]=0;
        vis[j]=1;
        queue<int> q;
        q.push(j);
        while(!q.empty()){
            int u=q.front();
            q.pop();
            vis[u]=0;
            for(i=head[u]; ~i; i=edge[i].next){
                int v=edge[i].to;
                if(low[j][v]>low[j][u]+edge[i].w){
                    low[j][v]=low[j][u]+edge[i].w;
                    if(!vis[v]){
                        vis[v]=1;
                        q.push(v);
                    }
                }
            }
        }
    }
}
int main()
{
    int m,i,j,x,y,s1,t1,s2,t2,l1,l2;
    cin>>n>>m;
    memset(head,-1,sizeof(head));
    memset(low,inf,sizeof(low));
    for(i=1;i<=m;++i){
        scanf("%d%d",&x,&y);
        add(x,y,1);
        add(y,x,1);
    }
    spfa();
    scanf("%d%d%d%d%d%d",&s1,&t1,&l1,&s2,&t2,&l2);
    if(low[s1][t1]>l1||low[s2][t2]>l2) puts("-1");
    else{
        int ans=low[s1][t1]+low[s2][t2];
        for(i=1;i<=n;++i)
            for(j=1;j<=n;++j)
            {
                if(low[s1][i]+low[i][j]+low[j][t1]<=l1&&low[s2][i]+low[i][j]+low[j][t2]<=l2)
                    ans=min(ans,low[s1][i]+low[i][j]+low[j][t1]+low[s2][i]+low[j][t2]);
                if(low[s1][i]+low[i][j]+low[j][t1]<=l1&&low[t2][i]+low[i][j]+low[j][s2]<=l2)
                    ans=min(ans,low[s1][i]+low[j][t1]+low[t2][i]+low[i][j]+low[j][s2]);
                
            }
        printf("%d\n",m-ans);
    }
    return 0;
}

 

The end；