Fzu 2271 X【Floyd】

 Problem 2271 X

Accept: 19    Submit: 80
Time Limit: 1500 mSec    Memory Limit : 32768 KB

 Problem Description

X is a fully prosperous country, especially known for its complicated transportation networks. But recently, for the sake of better controlling by the government, the president Fat Brother thinks it’s time to close some roads in order to make the transportation system more effective.

Country X has N cities, the cities are connected by some undirected roads and it’s possible to travel from one city to any other city by these roads. Now the president Fat Brother wants to know that how many roads can be closed at most such that the distance between any two cities in country X does not change. Note that the distance between city A and city B is the minimum total length of the roads you need to travel from A to B.

 Input

The first line of the date is an integer T (1 <= T <= 50), which is the number of the text cases.

Then T cases follow, each case starts with two numbers N, M (1 <= N <= 100, 1 <= M <= 40000) which describe the number of the cities and the number of the roads in country X. Each case goes with M lines, each line consists of three integers x, y, s (1 <= x, y <= N, 1 <= s <= 10, x is not equal to y), which means that there is a road between city x and city y and the length of it is s. Note that there may be more than one roads between two cities.

 Output

For each case, output the case number first, then output the number of the roads that could be closed. This number should be as large as possible.

See the sample input and output for more details.

 Sample Input

2

2 3

1 2 1

1 2 1

1 2 2

3 3

1 2 1

2 3 1

1 3 1

 Sample Output

Case 1: 2

Case 2: 0

题目大意：

让你删除尽可能多的边，使得剩下的边构成的图两点间最短距离不变。

思路：

我们要的是两点间最短距离不变，所以我们当存在：mp【j】【i】+mp【i】【k】<mp【j】【k】的时候，如果从j到k这条边是原图中的直接相连的边，那么这条边就可以删除。

那么我们设定vis【i】【j】；

vis【i】【j】=0表示从i到j还没有路。

vis【i】【j】=1表示从i到j的路是原来图中直接相连的边。

vis【i】【j】=2表示从i到j的路是通过松弛得到的。

那么对于存在：mp【j】【i】+mp【i】【k】<mp【j】【k】的时候，如果vis【j】【k】==1.那么就统计一次答案即可。

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;int Dist[150][150];int vis[150][150];int main(){    int t;    int kase=0;    scanf("%d",&t);    while(t--)    {        int n,m;        scanf("%d%d",&n,&m);        int output=0;        memset(vis,0,sizeof(vis));        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                Dist[i][j]=0x3f3f3f3f;            }        }        for(int i=1;i<=m;i++)        {            int x,y,w;            scanf("%d%d%d",&x,&y,&w);            if(0x3f3f3f3f==Dist[x][y])            {                vis[x][y]=1;                vis[y][x]=1;                Dist[x][y]=w;                Dist[y][x]=w;            }            else            {                if(w<Dist[x][y])                {                    Dist[x][y]=w;                    Dist[y][x]=w;                }                output++;            }        }        int cnt=0;        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                for(int k=1;k<=n;k++)                {                    if(vis[j][i]==0||vis[i][k]==0)continue;                    if(vis[j][k]==0)                    {                        Dist[j][k]=Dist[j][i]+Dist[i][k];                        vis[j][k]=2;                    }                    else if(vis[j][k]==1)                    {                        if(Dist[j][i]+Dist[i][k]<=Dist[j][k])                        {                            Dist[j][k]=Dist[j][i]+Dist[i][k];                            vis[j][k]=2;                            cnt++;                        }                    }                    else if(vis[j][k]==2)                    {                        if(Dist[j][i]+Dist[i][k]<Dist[j][k])                        {                            Dist[j][k]=Dist[j][i]+Dist[i][k];                        }                    }                }            }        }        printf("Case %d: %d\n",++kase,cnt/2+output);    }}