zoj 2760 How Many Shortest Path //MAXFLOW
How Many Shortest Path
Time Limit: 10 Seconds Memory Limit: 32768 KBGiven a weighted directed graph, we define the shortest path as the path who has the smallest length among all the path connecting the source vertex to the target vertex. And if two path is said to be non-overlapping, it means that the two path has no common edge. So, given a weighted directed graph, a source vertex and a target vertex, we are interested in how many non-overlapping shortest path could we find out at most.
Input
Input consists of multiple test cases. The first line of each test case, there is an integer number N (1<=N<=100), which is the number of the vertices. Then follows an N * N matrix, represents the directed graph. Each element of the matrix is either non-negative integer, denotes the length of the edge, or -1, which means there is no edge. At the last, the test case ends with two integer numbers S and T (0<=S, T<=N-1), that is, the starting and ending points. Process to the end of the file.
Output
For each test case, output one line, the number of the the non-overlapping shortest path that we can find at most, or "inf" (without quote), if the starting point meets with the ending.
Sample Input
4 0 1 1 -1 -1 0 1 1 -1 -1 0 1 -1 -1 -1 0 0 3 5 0 1 1 -1 -1 -1 0 1 1 -1 -1 -1 0 1 -1 -1 -1 -1 0 1 -1 -1 -1 -1 0 0 4
Sample Output
2 1
Author: SHEN, Guanghao
Source: ZOJ Monthly, September 2006
这道题学到不少东西。
以前做过求S-->T中的路中最长边最小的路径的数目这一类问题，当时的解法是二分+网络流
这道题目是求最短路的数量，竟然也可以用网络流做
首先以前没想到的，可以用FLOYED去算出I-->J这条边是否在最短路上，接着若是在则标记为1
注意，FLOYED的过程容易错，并且不要改变原来图中变量，因为可能原本I->J是没有边的，但你FLOYED后可能I-->J就有边了，你可能把I-->J这条边加入，那就错了
接着就可以用求最大流做了
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int inf=100000000;
int n,s,t;
int flow[110][110],mat[110][110],gap[110],dist[110];
int map[110][110];
int find_path(int p,int limit=1<<30-1)
{
if(p==n-1) return limit;
for(int i=0;i<n;i++)
{
if(dist[p]==dist[i]+1&&map[p][i]>0)
{
int t=find_path(i,min(limit,map[p][i]));
if(t<0) return t;
if(t>0)
{
map[p][i]-=t;
map[i][p]+=t;
return t;
}
}
}
int label=n;
for(int i=0;i<n;i++) if(map[p][i]>0) label=min(label,dist[i]+1);
if(--gap[dist[p]]==0||dist[0]>=n) return -1;
++gap[dist[p]=label];
return 0;
}
int iSAP()
{
gap[0]=n;
int maxflow=0,t=0;
while((t=find_path(0))>=0) maxflow+=t;
return maxflow;
}
void buildgraph()
{
memcpy(mat,flow,sizeof(flow));
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++)
{
if(mat[i][k]==inf) continue;
for(int j=1;j<=n;j++)
{
if(mat[k][j]==inf) continue;
mat[i][j]=min(mat[i][j],mat[i][k]+mat[k][j]);
}
}
memset(map,0,sizeof(map));
for(int i=1;i<=n;i++)
{
if(mat[s][i]==inf) continue;
for(int j=1;j<=n;j++)
{
if(mat[j][t]==inf) continue;
map[i][j]=(flow[i][j]!=inf && mat[s][i]+flow[i][j]+mat[j][t]==mat[s][t]);
}
}
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
memset(flow,0,sizeof(flow));
memset(gap,0,sizeof(gap));
memset(dist,0,sizeof(dist));
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
scanf("%d",&flow[i][j]);
if(i==j) flow[i][j]=0;
else if(flow[i][j]==-1) flow[i][j]=inf;
}
scanf("%d%d",&s,&t);
s++;t++;
if(s==t) printf("inf/n");
else
{
buildgraph();
map[0][s]=inf;
map[t][n+1]=inf;
n=n+2;
printf("%d/n",iSAP());
}
}
return 0;
}