题目地址:HDU 3435
这题刚上来一看,感觉毫无头绪。。再仔细想想。。发现跟我做的前两道费用流的题是差不多的。可以往那上面转换。
建图基本差不多,只不过这里是无向图。建图依然是拆点,判断入度出度,最后判断是否满流,满流的话这时的费用流是符合要求的,输出,不能满流的话,输出NO。
代码如下:
#include <iostream> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <math.h> #include <ctype.h> #include <queue> #include <map> #include<algorithm> using namespace std; const int INF=0x3f3f3f3f; int head[3000], source, sink, cnt, flow, cost, num; int d[3000], vis[3000], pre[3000], cur[3000]; queue<int>q; struct node { int u, v, cap, cost, next; }edge[10000000]; void add(int u, int v, int cap, int cost) { edge[cnt].v=v; edge[cnt].cap=cap; edge[cnt].cost=cost; edge[cnt].next=head[u]; head[u]=cnt++; edge[cnt].v=u; edge[cnt].cap=0; edge[cnt].cost=-cost; edge[cnt].next=head[v]; head[v]=cnt++; } int spfa() { memset(d,INF,sizeof(d)); memset(vis,0,sizeof(vis)); int minflow=INF, i; q.push(source); d[source]=0; cur[source]=-1; while(!q.empty()) { int u=q.front(); q.pop(); vis[u]=0; for(i=head[u];i!=-1;i=edge[i].next) { int v=edge[i].v; if(d[v]>d[u]+edge[i].cost&&edge[i].cap) { d[v]=d[u]+edge[i].cost; minflow=min(minflow,edge[i].cap); cur[v]=i; if(!vis[v]) { q.push(v); vis[v]=1; } } } } if(d[sink]==INF) return 0; flow+=minflow; cost+=minflow*d[sink]; for(i=cur[sink];i!=-1;i=cur[edge[i^1].v]) { edge[i].cap-=minflow; edge[i^1].cap+=minflow; } return 1; } void mcmf(int n) { while(spfa()); if(flow==n) printf("%d\n",cost); else printf("NO\n"); } int main() { int t, n, m, i, j, a, b, c; scanf("%d",&t); num=0; while(t--) { num++; scanf("%d%d",&n,&m); memset(head,-1,sizeof(head)); cnt=0; source=0; sink=2*n+1; flow=0; cost=0; for(i=1;i<=n;i++) { add(source,i,1,0); add(i+n,sink,1,0); } while(m--) { scanf("%d%d%d",&a,&b,&c); add(a,b+n,1,c); add(b,a+n,1,c); } printf("Case %d: ",num); mcmf(n); } return 0; }