Problem 2271 X （flyod变形）

Problem 2271 X
Accept: 113 Submit: 385
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
题意：在已知的图上，保证最短路不变的情况下，尽可能多的去掉一些边。因为点只有100个，而且要求所有点的最短路，所以用floyd，先求出任意两点之间的最短路，然后与原路比较，只要小的话就说明原路中这条路可以被替换掉，则可以删去。因为两点之间有多条边，所以要注意。还有就是如果原来的路mapp[i][j]等于最短路map[i][k]+map[k][j]，那么mapp[i][j]这条路也要被删除。

#include<stdio.h>#include<string.h>#include<algorithm>#define inf 0x3f3f3f3fusing namespace std;const int maxn=105;int n,m;int mapp[maxn][maxn];int map1[maxn][maxn];int main(){    int t,ans=1;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        int sum=0;        for(int i=1; i<=n; i++)            for(int j=1; j<=n; j++)            {                if(i==j)mapp[i][j]=0;                else mapp[i][j]=inf;            }        int a,b,c;        for(int i=1; i<=m; i++)        {            scanf("%d%d%d",&a,&b,&c);            if(mapp[a][b]==inf)                mapp[a][b]=c;            else            {                mapp[a][b]=min(mapp[a][b],c);                sum++;            }            mapp[b][a]=mapp[a][b];        }        for(int i=1; i<=n; i++)        {            for(int j=1; j<=n; j++)            {                map1[i][j]=mapp[i][j];            }        }        for(int k=1; k<=n; k++)            for(int i=1; i<=n; i++)                for(int j=1; j<=n; j++)                {                    if(map1[i][j]>map1[i][k]+map1[k][j])                        map1[i][j]=map1[i][k]+map1[k][j];                }        for(int i=1; i<=n; i++)            for(int j=i+1; j<=n; j++)            {                if(mapp[i][j]==inf)continue;                if(map1[i][j]<mapp[i][j])sum++;                else                {                    for(int k=1; k<=n; k++)                    {                        if(k==i||k==j)continue;                        else                        {                            if(mapp[i][j]==map1[i][k]+map1[k][j])                            {                                sum++;                                break;                            }                        }                    }                }            }            printf("Case %d: %d\n",ans++,sum);    }}