题目地址:NEU 1458
跟杭电上的那两个方格取数不太一样。。这个可以重复,但是取和的时候只能加一次。建图思路基本一会就出来。同样的拆点,只不过这题需要再拆个边,其中一条费用0,另一条费用为那个点处的值。流量都限制为1.然后剩下的都跟杭电上的那两个差不多了。因为把数组开小了WA了好几发。。(我前面居然还专门检查了一下数组大小,居然当时还认为没开小。。。对自己无语。。)
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[21000], source, sink, cnt, mp[101][101], flow, cost, n; int vis[21000], d[21000], cur[21000]; struct node { int u, v, cap, cost, next; } edge[1000000]; 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() { int i, minflow=INF; memset(d,INF,sizeof(d)); memset(vis,0,sizeof(vis)); cur[source]=-1; d[source]=0; queue<int>q; q.push(source); 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]) { vis[v]=1; q.push(v); } } } } if(d[sink]==INF) return 0; flow+=minflow; cost-=minflow*d[sink]; //printf("%d %d\n",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() { flow=cost=0; while(spfa()) ; printf("%d\n",cost); } int main() { int i, j; while(scanf("%d",&n)!=EOF) { for(i=1; i<=n; i++) { for(j=1; j<=n; j++) { scanf("%d",&mp[i][j]); } } memset(head,-1,sizeof(head)); cnt=0; source=0; sink=2*n*n+1; add(source,1,2,0); add(2*n*n,sink,2,0); for(i=1; i<=n; i++) { for(j=1; j<=n; j++) { add((i-1)*n+j,(i-1)*n+j+n*n,1,0); add((i-1)*n+j,(i-1)*n+j+n*n,1,-mp[i][j]); if(1+j<=n) { add((i-1)*n+j+n*n,(i-1)*n+j+1,2,0); } if(i+1<=n) { add((i-1)*n+j+n*n,i*n+j,2,0); } } } mcmf(); } return 0; }
NEU 1458 方格取数(网络流之费用流),布布扣,bubuko.com
原文地址:http://blog.csdn.net/scf0920/article/details/38362725