http://poj.org/problem?id=2195

題意是：有相同數(shù)量的人與房子，每一時(shí)刻人都可以花費(fèi)1$的錢走一步，問讓每個(gè)人到達(dá)一個(gè)屋子的最少需要的費(fèi)用。

建立源點(diǎn)與匯點(diǎn)，求有源點(diǎn)到匯點(diǎn)的最小費(fèi)用最大流；改了一下不需要f[][]的模板。

#include <iostream> #include <cstring> #include <cstdio> #include <queue> #include <cstdlib> #define maxn 207 using namespace std;const int inf = 999999999;struct node {int x,y; }p[maxn],h[maxn]; int c[maxn][maxn],f[maxn][maxn],w[maxn][maxn]; int pre[maxn],dis[maxn]; int n,m,pn,hn,s,t; char str[maxn][maxn]; bool inq[maxn]; int ans;int Abs(int x) {return x > 0 ? x : -x; } void spfa() {int v;queue<int>q;for (int i = 0; i < maxn; ++i){dis[i] = inf; pre[i] = -1;inq[i] = false;}q.push(s); inq[s] = true; dis[s] = 0;while (!q.empty()){int u = q.front(); q.pop();inq[u] = false;for (v = 0; v <= t; ++v){if (c[u][v]&& dis[v] > dis[u] + w[u][v]){dis[v] = dis[u] + w[u][v];pre[v] = u;if (!inq[v]){q.push(v);inq[v] = true;}}}} } void mcmf() {while (1){spfa();if (pre[t] == -1) break;int x = t,minf = inf;while (pre[x] != -1){minf = min(minf,c[pre[x]][x]);x = pre[x];}x = t;while (pre[x] != -1){c[pre[x]][x] -= minf;c[x][pre[x]] += minf;ans += minf*w[pre[x]][x];x = pre[x];}} } int main() {//freopen("in.txt","r",stdin);int i,j,pn,hn;while (~scanf("%d%d",&n,&m)){if (!n && !m) break;pn = hn = 0;for (i = 0; i < n; ++i){scanf("%s",str[i]);for (j = 0; j < m; ++j){if (str[i][j] == 'H'){h[++hn].x = i; h[hn].y = j;}else if (str[i][j] == 'm'){p[++pn].x = i; p[pn].y = j;}}}memset(c,0,sizeof(c));memset(f,0,sizeof(f));memset(w,0,sizeof(w));s = 0; t = pn + hn + 1;for (i = 1; i <= pn; ++i) c[s][i] = 1;for (i = 1; i <= hn; ++i) c[i + pn][t] = 1;for (i = 1; i <= pn; ++i){for (j = 1; j <= hn; ++j){c[i][j + pn] = 1;w[i][j + pn] = Abs(p[i].x - h[j].x) + Abs(p[i].y - h[j].y);w[j + pn][i] = -w[i][j + pn];}}ans = 0;mcmf();printf("%d\n",ans);}return 0; }

　　

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