1059: [ZJOI2007]矩阵游戏

1059: [ZJOI2007]矩阵游戏

链接

思路：

　　可以在对角线上填满的条件是，每一行，每一列，都要有一个数字。所以把列看成一排点，行看成一排点。二分图匹配，要求匹配数是n，即全匹配成功。

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cctype>
 
using namespace std;

const int N = 510; // 两倍空间！ 
const int INF = 1e9;
struct Edge{
    int to,nxt,c;
    Edge() {}
    Edge(int x,int y,int z) {to = x;c = y;nxt = z;}
}e[100100];
int head[N],dis[N],cur[N],q[100100];
int S,T,tot,L,R;

inline int read() {
    int x = 0,f = 1;char ch = getchar();
    for (; !isdigit(ch); ch=getchar()) if(ch=='-') f = -1;
    for (; isdigit(ch); ch=getchar()) x = x*10+ch-'0';
    return x * f;
}

void init() {
    tot = 1;
    memset(head,0,sizeof(head));
}
void add_edge(int u,int v,int w) {
    e[++tot] = Edge(v,w,head[u]);head[u] = tot;
    e[++tot] = Edge(u,0,head[v]);head[v] = tot;
}
bool bfs() {
    for (int i=1; i<=T; ++i) cur[i] = head[i],dis[i] = -1;
    L = 1;R = 0;
    q[++R] = S;
    dis[S] = 1;
    while (L <= R) {
        int u = q[L++];
        for (int i=head[u]; i; i=e[i].nxt) {
            int v = e[i].to;
            if (dis[v] == -1 && e[i].c > 0) {
                dis[v] = dis[u] + 1;
                q[++R] = v;
                if (v==T) return true;
            }
        }
    }
    return false;
}
int dfs(int u,int flow) {
    if (u == T) return flow;
    int used = 0,tmp;
    for (int &i=cur[u]; i; i=e[i].nxt) {
        int v = e[i].to;
        if (dis[v] == dis[u] + 1 && e[i].c > 0) {
            tmp = dfs(v,min(e[i].c,flow-used));
            if (tmp) {
                e[i].c -= tmp;e[i^1].c += tmp;
                used += tmp;
                if (used == flow) break;
            }
        }
    }
    if (used != flow) dis[u] = -1;
    return used;
}
int dinic() {
    int ans = 0;
    while (bfs()) ans += dfs(S,INF);
    return ans;
}
int main() {
    int Case = read(); // 不要设成T 
    while (Case--) {
        init();
        int n = read();
        S = n+n+1,T = n+n+2;
        for (int i=1; i<=n; ++i) 
            for (int j=1; j<=n; ++j) {
                int a = read();
                if (a) add_edge(i,j+n,1);
            }
        for (int i=1; i<=n; ++i) 
            add_edge(S,i,1),add_edge(i+n,T,1);
        int ans = dinic();
        if (ans==n) puts("Yes");
        else puts("No");
    }
    return 0;
}