裸floyd大法

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这个题主要就是裸的floyd大法注意就是不要计算了重边所以必须用vis标记...orz

C - D

Time Limit:1000MS Memory Limit:65768KB 64bit IO Format:%I64d & %I64u

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Description

Everyone knows how to calculate the shortest path in a directed graph. In fact, the opposite problem is also easy. Given the length of shortest path between each pair of vertexes, can you find the original graph?

Input

The first line is the test case number T (T ≤ 100).
First line of each case is an integer N (1 ≤ N ≤ 100), the number of vertexes.
Following N lines each contains N integers. All these integers are less than 1000000.
The jth integer of ith line is the shortest path from vertex i to j.
The ith element of ith line is always 0. Other elements are all positive.

Output

For each case, you should output “Case k: ” first, where k indicates the case number and counts from one. Then one integer, the minimum possible edge number in original graph. Output “impossible” if such graph doesn't exist.

Sample Input

 330 1 11 0 11 1 030 1 3 4 0 27 3 030 1 41 0 24 2 0

Sample Output

 Case 1: 6Case 2: 4Case 3: impossible

#include<cstdio>int line[101][101];int ver;int floyd(){    int cnt=0;    bool vis[101][101]={0};    for(int k=1;k<=ver;k++){        for(int i=0;i<=ver;i++){            for(int j=0;j<=ver;j++){                if(i==j||j==k||k==i)continue;                if(!vis[i][j]&&line[i][j]==line[i][k]+line[k][j]){                    vis[i][j]=1;                    cnt++;                }                if(line[i][j]>line[i][k]+line[k][j])return -1;            }        }    }    return cnt;}int main(){    int cas,i,j,k,ans;    scanf("%d",&cas);    for(i=0;i<cas;i++){        scanf("%d",&ver);        for(j=1;j<=ver;j++){            for(k=1;k<=ver;k++)scanf(" %d",&line[j][k]);        }        ans=floyd();        if(ans==-1)printf("Case %d: impossible\n",i+1);        else {            printf("Case %d: %d\n",i+1,ver*(ver-1)-ans);        }    }    return 0;}

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