POJ 3216 Folyd +最小路径覆盖

先由Floyd求出2点间最短距离，再由任务之间的关系来构图

#include <stdio.h>#include <string.h>#define N 201#define INF 0x7ffffffbool map[N][N];int link[N];bool use[N];int g[21][21];struct Job{int city;int st;//start timeint d;//time cost}t[N];int n,q;int min (int a,int b){return a>b? b:a;}void floyd(){int i,j,k;for (k=1;k<=q;k++){for(i=1;i<=q;i++){for (j=1;j<=q;j++){g[i][j]=min(g[i][j],g[i][k]+g[k][j]);}}}}int search_path(int i){int v;for (v=1;v<=n;v++){if (!use[v] && map[i][v]){use[v]=1;if (link[v]==-1 || search_path(link[v])){link[v]=i;return 1;}}}return 0;}int max_match(){int sum=0,i;memset(link,-1,sizeof(link));for (i=1;i<=n;i++){memset(use,0,sizeof(use));if (search_path(i)) sum++;}return sum;}int main (){//freopen("3216.txt","r",stdin);int i,j,a,b,sum;int pt;//pretime 前一个工作要的时间while(scanf("%d%d",&q,&n),q||n){memset(map,0,sizeof(map));for (i=1;i<=q;i++){for (j=1;j<=q;j++){scanf("%d",&g[i][j]);if (g[i][j]==-1) g[i][j]=INF;}}for (i=1;i<=n;i++)scanf("%d%d%d",&t[i].city,&t[i].st,&t[i].d);floyd();for (i=1;i<=n;i++){pt=t[i].st+t[i].d;for (j=1;j<=n;j++){if (i==j) continue;a=t[i].city;b=t[j].city;if (g[a][b]+pt<=t[j].st) map[i][j]=1; //做完i 还能赶上j 就在i j工作间连上线}}sum=max_match();printf("%d/n",n-sum);}}