HDU 3395 Special Fish

以前通过KM来解的，今天用最大费用最大流写一次，结果发现WA。。次奥，这不科学。。。后来才发现有一种情况没有考虑到，即首先满足的应该是最大费用，而不是流。。。

这儿有两种解决方法：

1、一种是在加边，即把二分图左边的鱼与汇点相连，流量为1，费用为0；这样才能保证在最大费用时，流量也是最大的。传送门

2、直接做最小费用流（不是最小费用最大流），将增广的结束条件改为d[t]>=0即可。如果网络负费用圈，则需要消圈。

#include <iostream>#include <cstdlib>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <vector>#include <queue>#include <algorithm>#include <map>using namespace std;const int maxn = 1010;const int INF = 0x3f3f3f3f;struct Edge{int from, to, cap, flow, cost;Edge(int from, int to, int cap, int flow, int cost): from(from), to(to), cap(cap), flow(flow), cost(cost) {}};struct MCMF{int n, m, s, t;vector<Edge> edges;vector<int> G[maxn];int inq[maxn];int d[maxn];int p[maxn];int a[maxn];void init(int n){this->n = n;for(int i = 0; i <= n; i++) G[i].clear();edges.clear();}void AddEdge(int from, int to, int cap, int cost){edges.push_back(Edge (from, to, cap, 0, cost));edges.push_back(Edge (to, from, 0, 0, -cost));m = edges.size();G[from].push_back(m-2);G[to].push_back(m-1);}bool spfa(int s, int t, int &flow, int &cost){for(int i = 0; i <= n; i++) d[i] = INF;memset(inq, 0, sizeof(inq));d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;queue<int> Q;Q.push(s);while(!Q.empty()){int u = Q.front(); Q.pop();inq[u] = 0;for(int i = 0; i < G[u].size(); i++){Edge &e = edges[G[u][i]];if(e.cap > e.flow && d[e.to] > d[u]+e.cost){d[e.to] = d[u]+e.cost;p[e.to] = G[u][i];a[e.to] = min(a[u], e.cap-e.flow);if(!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }}}}if(d[t] == INF) return 0;flow += a[t];cost += d[t]*a[t];int u = t;while(u != s){edges[p[u]].flow += a[t];edges[p[u]^1].flow -= a[t];u = edges[p[u]].from;}return 1;}int Mincost(int s, int t, int &flow, int &cost){while(spfa(s, t, flow, cost));return cost;}};void readint(int &x){    char c;    while(!isdigit(c)) c = getchar();        x = 0;    while(isdigit(c))    {        x = x*10 + c-'0';        c = getchar();    }}void writeint(int x){    if(x > 9) writeint(x/10);    putchar(x%10+'0');}///////////////////////////////////////MCMF solver;int n, m, s, t;int C[maxn];char str[110][110];int main(){for(;;){readint(n);if(!n) break;for(int i = 1; i <= n; i++) scanf("%d", &C[i]);solver.init(2*n+10);s = 2*n+1, t = 2*n+2;for(int i = 1; i <= n; i++) scanf("%s", str[i]+1);for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++){if(str[i][j] == '1') solver.AddEdge(i, j+n, 1, -(C[i]^C[j]));}for(int i = 1; i <= n; i++){solver.AddEdge(s, i, 1, 0);solver.AddEdge(i+n, t, 1, 0);solver.AddEdge(i, t, 1, 0);}int cost = 0, flow = 0;solver.Mincost(s, t, flow, cost);printf("%d\n", -cost);}}