hdu 4411 Arrest 费用流模板

题意：警察局在0点，里面有k个警察，要将1～n的贼窝一网打尽，这1+n个点都有距离，且要求抓 i 点的贼前保证已经抓光 比i小的贼。警察们最后要回到0点，问满足抓到所以的贼（题目保证可行）最少走的路之和

建边参照yyn http://blog.csdn.net/u013368721/article/details/38781127

#include <bits/stdc++.h>#include <map>#include <set>#include <queue>#include <stack>#include <cmath>#include <time.h>#include <vector>#include <cstdio>#include <string>#include <iomanip>///cout << fixed << setprecision(13) << (double) x << endl;#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#define ls rt << 1#define rs rt << 1 | 1#define pi acos(-1.0)#define eps 1e-8#define Mp(a, b) make_pair(a, b)#define asd puts("asdasdasdasdasdf");typedef long long ll;//typedef __int64 LL;const int inf = 0x3f3f3f3f;const int M = 100010;const int N = 220;const int st = N-9, ed = N-8;struct node{int v, w, c, nxt;//c--->cap  w--->cost(dis)}e[M];int cnt, n, m, k;int head[N];int dis[N];//费用看成距离int cap[N];//流量int cur[N];//当前弧int vis[N];queue <int> q;int mat[N][N];int flow, cost;void init(){cnt = 0;memset( head, -1, sizeof( head ) );memset( mat, inf, sizeof( mat ) );}void add( int u, int v, int c, int w ){e[cnt].v = v;e[cnt].c = c;e[cnt].w = w;e[cnt].nxt = head[u];head[u] = cnt++;e[cnt].v = u;e[cnt].c = 0;e[cnt].w = -w;e[cnt].nxt = head[v];head[v] = cnt++;}bool spfa(){memset( dis, inf, sizeof( dis ) );memset( vis, 0, sizeof( vis ) );while( !q.empty() )q.pop();cap[st] = inf;cur[st] = -1;dis[st] = 0;q.push( st );while( !q.empty() ) {int u = q.front();q.pop();vis[u] = 0;for( int i = head[u]; ~i; i = e[i].nxt ) {int v = e[i].v, c = e[i].c, w = e[i].w;if( c && dis[v] > dis[u] + w ) {dis[v] = dis[u] + w;cap[v] = min( c, cap[u] );cur[v] = i;if( !vis[v] ) {vis[v] = 1;q.push(v);}}}}if( dis[ed] == inf )return 0;flow += cap[ed];cost += cap[ed] * dis[ed];for( int i = cur[ed]; ~i; i = cur[e[i^1].v] ) {e[i].c -= cap[ed];e[i^1].c += cap[ed];}return 1;}int MCMF(){flow = cost = 0;while( spfa() );return cost;}int main(){while( ~scanf("%d%d%d", &n, &m, &k) ) {if( n == m && n == k && n == 0 )break;init();for( int i = 1, u, v, w; i <= m; ++i ) {scanf("%d%d%d", &u, &v, &w);mat[u][v] = mat[v][u] = min( mat[u][v], w );}for( int k = 0; k <= n; ++k ) {for( int i = 0; i <= n; ++i ) {for( int j = 0; j <= n; ++j ) {mat[i][j] = min( mat[i][j], mat[i][k]+mat[k][j] );}}}for( int i = 1; i <= n; ++i )add( i, i+n, 1, -M );//保证一定要经过i点，但是导致费用为负，在最后加回来就行了。for( int i = 1; i <= n; ++i ) {add( 0, i, 1, mat[0][i] );//送一个警察去抓贼add( i+n, ed, 1, mat[i][0] );//回到家}for( int i = 1; i <= n; ++i ) {for( int j = i+1; j <= n; ++j )add( i+n, j, 1, mat[i][j] );}add( st, 0, k, 0 );add( 0, ed, k, 0 );int ans = MCMF() + n * M;//加回n*Mprintf("%d\n", ans);}return 0;}