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

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();
    }
}