hdu 2448 Mining Station on the Sea（最短路径+KM）

 

 

http://acm.hdu.edu.cn/showproblem.php?pid=2448

 

题目大意：海上有M个油田，有N个的船只和港口，N<=M，N艘船停在N个油田，油田和油田之间有距离，油田和港口之间也有距离。现在呢要你把船开会港口，每个港口只能停一艘船，问你怎么停可以使船只行驶的距离最短呢。

 

解题思路：先把船只所停的油田作为起点，做N次的最短路径，分别求出，从源点到港口的最短路径，这个就作为它们的权值，之后KM算法求最小权匹配就好。

 

#include <stdio.h>#include <iostream>#include <string.h>#include <queue>#define Max 305#define inf 99999999using namespace std;int n,m,k,p;bool visitedX[Max],visitedY[Max];int lx[Max],ly[Max];int Match[Max];int slack[Max];int Graph[Max][Max];int o[Max];/*源点*/struct node{int adj[Max],w[Max];int len;}list[Max];/*添加结点s的邻接点信息*/void AddAdj(int s,int e,int w){list[s].adj[list[s].len] = e;list[s].w[list[s].len++] = w;}void SPFA(int locate){int dist[Max],i,u,v,w;bool visited[Max];queue<int>q;for(i=1;i<=n+m;i++)visited[i] = 0,dist[i] = inf;visited[locate] = 1;dist[locate] = 0;q.push(locate);while (!q.empty()){u = q.front();q.pop();visited[u] = 0;for (i=0;i<list[u].len;i++){v = list[u].adj[i];w = list[u].w[i];if (dist[v]>dist[u]+w){dist[v] = dist[u]+w;if(!visited[v])q.push(v),visited[v] = 1;}}}/*构建源点到港口最小权值图*/for (i=m+1;i<=m+n;i++){if(dist[i]!=0x7fffffff)Graph[locate][i] = -dist[i];/*因为是求最小权值匹配，所以权值取反*/elseGraph[locate][i] = 0;}}bool DFS(int i){int j,tem;visitedX[i] = true;/*港口下标从m+1开始，m为油田数量*/for (j=m+1;j<=n+m;j++){if (!visitedY[j]){tem = lx[i]+ly[j]-Graph[i][j];if (tem==0){visitedY[j] = 1;if (Match[j]==-1||DFS(Match[j])){Match[j] = i;return true;}}else if(slack[j]>tem)slack[j] = tem;}}return false;}/*二分图最优权匹配*/void KM(){int i,j,k,min;for (i=1;i<=n;i++){/*X集合中的元素为o[i]*/lx[o[i]] = -inf;ly[i+m] = 0;Match[i+m] = -1;/*j指向Y集合下标，也就是港口的下标*/for (j=m+1;j<=m+n;j++)if(lx[o[i]]<Graph[o[i]][j])lx[o[i]] = Graph[o[i]][j];}for (k=1;k<=n;k++){for(i=m+1;i<=n+m;i++)slack[i] = inf;while (1){min = inf;memset(visitedX,0,sizeof(visitedX));memset(visitedY,0,sizeof(visitedY));if(DFS(o[k]))break;for(i=1+m;i<=n+m;i++){if(!visitedY[i]&&slack[i]<min)min = slack[i];}/*如果找不到路径，就取出所有被访问过的X集合和Y集合中的点对起标量处理*/for(i=1;i<=n;i++){if(visitedX[o[i]])lx[o[i]]-=min;if(visitedY[i+m])ly[i+m]+=min;elseslack[i+m]-=min;}}}}int main(){int i,j,val,sum;while (scanf("%d%d%d%d",&n,&m,&k,&p)!=EOF){for(i=1;i<=n;i++)scanf("%d",&o[i]);for(i=1;i<=m;i++)list[i].len = 0;while (k--){scanf("%d%d%d",&i,&j,&val);/*油田之间路是双向的*/AddAdj(i,j,val);AddAdj(j,i,val);}while (p--){scanf("%d%d%d",&i,&j,&val);/*油田和港口之间路是单向的，避免求最短路径时，港口作为中继点，求错路径*/AddAdj(j,i+m,val);}memset(Graph,0,sizeof(Graph));/*分别对n个船只所在点作为起点，求到各港口的最短路径*/for(i=1;i<=n;i++)SPFA(o[i]);KM();for(sum = 0,i=m+1;i<=n+m;i++)sum+=lx[Match[i]]+ly[i];printf("%d/n",-sum);}return 0;}