hdu 4322 Candy 【最大费用最大流】

Problem:

有n个糖果和m个孩子，给出每个孩子是否喜欢某个糖果，如果把孩子孩子喜欢的糖果给了孩子，那个孩子会增加k个glad值，否则只增加一个，问能否有一个方案使得所有的孩子的glad值都不小于b[i]?

Solution:

用最大费用最大流在能使glad值最大的情况下求出最优匹配的糖果个数，然后看剩下的glad值能否由剩下的糖果满足。

超级源点和糖果间容量为1，费用为0.

孩子和喜欢的糖果之间容量为1，费用为0。

孩子和汇点之间容量为b[i]/k，费用为k。

如果b[i]不能整除k，则再添加一条边容量为1，费用为b[i]%k。

note:

1. 有负权边要使用spfa，保存图可以使用rev和revv和reve。

2. 提前建立一条容量为0的反向边以方便反向查找。

#include<cstdio>
#include<iostream>
#include<sstream>
#include<cstdlib>
#include<cmath>
#include<cctype>
#include<string>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<ctime>
#include<vector>
#include<fstream>
#include<list>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
#define ms(s) memset(s,0,sizeof(s))

const double PI = 3.141592653589;
const int INF = 0x3fffffff;

struct Edge {
int to, cap, cost, rev;
Edge(int too, int capp, int costt, int revv) : to(too), cap(capp), cost(costt), rev(revv) {}
};

vector<vector<Edge> > G;
int st, en;
int dist[30];
int prevv[30], preve[30];

void add_edge(int from, int to, int cap, int cost) {
G[from].push_back(Edge(to, cap, cost, G[to].size()));
G[to].push_back(Edge(from, 0, -cost, G[from].size()-1));
}

int min_max_flow(int n) {//the quantity of points.
int ans = 0;//return the greatest flow.
while(true) {//spfa
fill(dist, dist+n, INF);
dist[st] = 0;
queue<int> q;  q.push(st);

while(!q.empty()) {
int u = q.front();  q.pop();
for(int i = 0; i < G[u].size(); i++) {
int to = G[u][i].to;
if(G[u][i].cap > 0 && dist[u] + G[u][i].cost < dist[to]) {
dist[to] = dist[u] + G[u][i].cost;
prevv[to] = u;  preve[to] = i;
q.push(to);
}
}
}
if(dist[en] == INF)
break;
else {
int f = INF;
for(int pr = en; pr != st; pr = prevv[pr])
f = min(f, G[prevv[pr]][preve[pr]].cap);
ans += f;
for(int pr = en; pr != st; pr = prevv[pr]) {
Edge &e = G[prevv[pr]][preve[pr]];
e.cap -= f;
G[pr][e.rev].cap += f;
}
}
}
return ans;
}

int main() {
//        freopen("/Users/really/Documents/code/input","r",stdin);
//    freopen("/Users/really/Documents/code/output","w",stdout);
ios::sync_with_stdio(false);

int t;
cin >> t;
for(int idx = 1; idx <= t; idx++) {
G.clear();
int b[20];
int n, m, k;
cin >> n >> m >> k;
st = 0;  en = n+m+1;  G.resize(n+m+2);
for(int i = 1; i <= m; i++)
cin >> b[i];
for(int i = 1; i <= n; i++)//connect start point to each candy.
add_edge(st, i, 1, 0);
for(int i = 1; i <= m; i++) {
for(int j = 1; j <= n; j++) {
int t;
cin >> t;
if(t == 1)
add_edge(j, n+i, 1, 0);
}
}
for(int i = 1; i <= m; i++) {
add_edge(n+i, en, b[i]/k, -k);
if(b[i]%k != 0)
add_edge(n+i, en, 1, -(b[i]%k));
}

int now = min_max_flow(n+m+2);
int res = 0;
for(int i = 1; i <= m; i++) {
for(int j = 0; j < G[n+i].size(); j++) {
if(G[n+i][j].to == en) {
if(G[n+i][j].cap > 0)
res -= G[n+i][j].cap*G[n+i][j].cost;
}
}
}
if(n-now >= res)
cout << "Case #" << idx << ": YES" << endl;
else
cout << "Case #" << idx << ": NO" << endl;
}

return 0;
}