hdu 4411 Arrest(最小费用最大流)

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4411

题目大意：从0开始按1到N的顺序，遍历所有的点，并回到0点；

解题思路：费用流！

#include<iostream>

#include<string.h>
#include<stdio.h>
#include<queue>
using namespace std;

#define MAXN 205
#define MAXE 16005
#define INF 1000000

queue<int> que;
int g[MAXN][MAXN];
struct{
   int node, cap, cost, next;
}edge[MAXE];
int ans, cnt, n;
int head[MAXN];
int pre[MAXN], dis[MAXN];
bool vis[MAXN];
void addEdge(int u, int node, int ca, int co){
   edge[cnt].node = node;
   edge[cnt].cap = ca;
   edge[cnt].cost = co;
   edge[cnt].next = head[u];
   head[u] = cnt ++;
   edge[cnt].node = u;
   edge[cnt].cap = 0;
   edge[cnt].cost = -co;
   edge[cnt].next = head[node];
   head[node] = cnt ++;
}
bool spfa(){                  // 源点为0，汇点为n。
   int i, u = 0;
   memset(vis, 0, sizeof(vis));
   while(!que.empty())que.pop();
   for(i = 1; i <= n; i++)dis[i] = INF;
   dis[0] = 0;
   que.push(0);
   vis[0] = true;
   while(!que.empty()){       // 这里最好用队列，有广搜的意思，堆栈像深搜。
       int u = que.front();
       que.pop();
       for(vis[u] = false,i = head[u]; ~i; i = edge[i].next){
           int node = edge[i].node;
           if(edge[i].cap && dis[node] > dis[u] + edge[i].cost){
               dis[node] = dis[u] + edge[i].cost;
               pre[node] = i;
               if(!vis[node]){
                   vis[node] = true;
                   que.push(node);
               }
           }
       }
   }
   if(dis[n] == INF) return false;
   return true;
}
void end(){
   int u, p, sum = INF;
   for(u = n; u != 0; u = edge[p^1].node){
       p = pre[u];
       sum = min(sum, edge[p].cap);
   }
   //maxflow += sum；//求最大流
   for(u = n; u != 0; u = edge[p^1].node){
       p = pre[u];
       edge[p].cap -= sum;
       edge[p^1].cap += sum;
       ans += sum * edge[p].cost;     // cost记录的为单位流量费用，必须得乘以流量。
   }
}
void floyd(int N)
{
   int i, j, k;
   for(k = 0; k <= N; k++)
       for(i = 0; i <= N; i++)
       {
           for(j = 0; j <= N; j++)
           {
               if(g[i][j] > g[i][k] + g[k][j])
                   g[i][j] = g[j][i] = g[i][k] + g[k][j];
           }
       }
       return ;
}
int main(){
   int N, M, K, i, j, s, t, a, b, c;
   while(scanf("%d%d%d", &N, &M, &K)&&(N + M + K))
   {
       for(i = 0; i <= N; i++)
       {
           g[i][i] = 0;
           for(j = i + 1; j <= N; ++ j )
               g[i][j] = g[j][i] = INF;
       }
       while(M--)
       {
           scanf("%d%d%d", &a, &b, &c);
           if(g[a][b]>c)g[a][b] = g[b][a] = c;
       }
       floyd(N);
       s = 0,t = 2 * N + 2;
       cnt = 0;
       memset(head, -1, sizeof(head));
       for(i = 0; i <= N; i++)
       {
           if(i)addEdge(i, N + i + 1, 1, -INF);
           else addEdge(i, N + i + 1, K, 0);
           for(j = i + 1; j <= N; ++ j )
               addEdge(N + i + 1, j, INF, g[i][j]);
       }
       for(i = 0; i <= N; i++)
           addEdge(i + N + 1, t, INF, g[0][i]);
       ans = 0;//ans为最小费用
       //maxflow = 0；
       n = t;
       while(spfa())
           end();
       printf("%d\n", ans + N * INF);
   }
   return 0;
}
/*
1 1 1
0 1 1

3 4 2
0 1 3
0 2 4
1 3 2
2 3 2
0 0 0
*/