UVa 10746 Crime Wava-The Sequel (最小费用最大流 + 精度）

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这道题我提交很多次，多事submission error 后来重新写了，就好了，总结一下，submission error 还是说明代码有问题，这次我开始的时候没有考虑到精度问题，而且数组也开得很小

做题的时候，切忌心烦意乱，要平稳一些，仔细一些，对于最大流的问题，要注意到点的数量，和标号，这个千万不要错

还有，最小费用最大流，一定要给每条边的反向边也赋初值，互为反向边的权值是互为倒数

代码如下：

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const double INFc = 1000000;const int INF = 1000000;const int N = 50;const double eps = 1e-9;int n, m, s, end;int cap[N][N], flow[N][N];double cost[N][N];void build() {    //源点和police相连    for ( int i = 1; i <= m; ++i ) cap[s][i] = 1, cost[s][i] = 0.0;    //bank和汇点相连    for ( int i = m+1; i < end; ++i ) cap[i][end] = 1, cost[i][end] = 0.0;    //police和bank相连    for ( int i = m+1; i < end; ++i )        for ( int j = 1; j <= m; ++j ) {            double c;            scanf("%lf", &c);            cost[j][i] = c;            cost[i][j] = -c;            cap[j][i] = 1;        }}void init() {    s = 0, end = n + m + 1;    memset( cap, 0, sizeof(cap));    memset( flow, 0, sizeof(flow));    for ( int i = 0; i <= end; ++i )         for ( int j = 0; j <= end; ++j ) cost[i][j] = INFc;}void mincost() {    queue<int> q;    int p[N], a;    bool vis[N];    double d[N], c = 0.0;    while ( 1 ) {        memset( vis, 0, sizeof(vis));        for ( int i = 0; i <= end; ++i ) d[i] = INFc;        d[s] = 0.0;        q.push(s);        while ( !q.empty() ) {            int u = q.front(); q.pop();            vis[u] = false;            for ( int v = s; v <= end; ++v )                 if ( cap[u][v] > flow[u][v] && d[v] > d[u] + cost[u][v] + eps ) {                    d[v] = d[u] + cost[u][v];                    p[v] = u;                    if ( !vis[v] ) {                        q.push(v);                        vis[v] = true;                    }                }        }        if ( d[end] == INFc ) break;        a = INF;        for ( int u = end; u != s; u = p[u] ) a = min( a, cap[p[u]][u] - flow[p[u]][u] );        for ( int u = end; u != s; u = p[u] ) {            flow[p[u]][u] += a;            flow[u][p[u]] -= a;        }        c += d[end] * a;    }    c /= n;    printf("%.2lf\n", c+eps);}int main(){    while ( scanf("%d%d", &n, &m) != EOF && !( !n && !m ) ) {        init();        build();        mincost();    }}