题目地址:HDU 2448
求n次最短路,将n艘船到各港口的最短路求出来,然后用最短路当费用,跑一次费用流。
代码如下:
#include <iostream> #include <cstdio> #include <string> #include <cstring> #include <stdlib.h> #include <math.h> #include <ctype.h> #include <queue> #include <map> #include <set> #include <algorithm> using namespace std; const int INF=0x3f3f3f3f; int head[400], cnt, source, sink, a[400], flow, cost; int d[400], f[400], vis[400], cur[400]; 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() { memset(vis,0,sizeof(vis)); memset(d,INF,sizeof(d)); f[source]=INF; 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(int 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; f[v]=min(f[u],edge[i].cap); cur[v]=i; if(!vis[v]) { vis[v]=1; q.push(v); } } } } if(d[sink]==INF) return 0; flow+=f[sink]; cost+=f[sink]*d[sink]; for(int i=cur[sink];i!=-1;i=cur[edge[i^1].v]) { edge[i].cap-=f[sink]; edge[i^1].cap+=f[sink]; } return 1; } void mcmf() { flow=cost=0; while(spfa()) ; printf("%d\n",cost); } int head1[400], cnt1, dp[400][400]; struct node1 { int u, v, w, next; }bian[400]; void add1(int u, int v, int w) { bian[cnt1].v=v; bian[cnt1].w=w; bian[cnt1].next=head1[u]; head1[u]=cnt1++; } void spfa(int x) { memset(vis,0,sizeof(vis)); memset(dp[x],INF,sizeof(dp[x])); dp[x][x]=0; queue<int>q; q.push(x); while(!q.empty()) { int u=q.front(); q.pop(); vis[u]=0; for(int i=head1[u];i!=-1;i=bian[i].next) { int v=bian[i].v; if(dp[x][v]>dp[x][u]+bian[i].w) { dp[x][v]=dp[x][u]+bian[i].w; if(!vis[v]) { vis[v]=1; q.push(v); } } } } } int main() { int n, m, k, p, i, j, u, v, w; while(scanf("%d%d%d%d",&n,&m,&k,&p)!=EOF) { memset(head,-1,sizeof(head)); memset(head1,-1,sizeof(head1)); cnt1=0; cnt=0; for(i=1;i<=n;i++) { scanf("%d",&a[i]); } while(k--) { scanf("%d%d%d",&u,&v,&w); add1(u,v,w); add1(v,u,w); } while(p--) { scanf("%d%d%d",&u,&v,&w); add1(v,u+m,w); } for(i=1;i<=n;i++) { spfa(a[i]); } source=0; sink=2*n+1; for(i=1;i<=n;i++) { add(source,i,1,0); add(i+n,sink,1,0); for(j=1+m;j<=m+n;j++) { if(dp[a[i]][j]!=INF) { add(i,j-m+n,1,dp[a[i]][j]); //printf("%d %d %d\n",i,j-m+n,dp[a[i]][j]); } } } mcmf(); } return 0; }
HDU 2448 Mining Station on the Sea(费用流)
原文地址:http://blog.csdn.net/scf0920/article/details/39692937