题目链接
划重点！！！每个有菜的点只能浇一次且恰好一次，所以意思就是，譬如某个菜的位置是(x, y)，那么，行x、列y的浇水方案只能使用其中的一个。
以此类推，我们给每个有蔬菜的位置的(x, y)的x点与y点链接一条无向边，代表x和y只能选择其中的一个，那么这个问题不就转化成了0、1染色问题了嘛？然后选择权值最少的0、1块就可以了。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define pii pair<int, int>
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e2 + 7;
const int maxP = maxN << 1, maxM = (maxN * maxN) << 1;
int N, M, mp[maxN][maxN], node, c[maxP];
namespace Graph
{int head[maxP], cnt;struct Eddge{int nex, to;Eddge(int a=-1, int b=0):nex(a), to(b) {}} edge[maxM];inline void addEddge(int u, int v){edge[cnt] = Eddge(head[u], v);head[u] = cnt++;}inline void _add(int u, int v) { addEddge(u, v); addEddge(v, u); }inline void init(){cnt = 0;for(int i=1; i<=node; i++) head[i] = -1;}
};
using namespace Graph;
int col[maxP], sum[2], que[maxP], top, tail;;
void bfs(int S)
{memset(sum, 0, sizeof(sum));top = tail = 0;col[S] = 0; sum[0] += c[S];que[tail++] = S;while(top < tail){int u = que[top++];for(int i=head[u], v; ~i; i=edge[i].nex){v = edge[i].to;if(~col[v]) continue;col[v] = col[u] ^ 1;sum[col[v]] += c[v];que[tail++] = v;}}
}
int main()
{int T; scanf("%d", &T);for(int Cas=1; Cas <= T; Cas++){scanf("%d%d", &N, &M);node = N + M; for(int i=1; i<=node; i++) col[i] = -1;init();for(int i=1; i<=N; i++) for(int j=1; j<=M; j++){scanf("%d", &mp[i][j]);if(mp[i][j]) _add(i, N + j);}for(int i=1; i<=node; i++) scanf("%d", &c[i]);int ans = 0;for(int i=1; i<=node; i++){if(~col[i]) continue;bfs(i);ans += min(sum[0], sum[1]);}printf("Case #%d: %d\n", Cas, ans);}return 0;
}