P1129 [ZJOI2007]矩阵游戏 【最大流】

 

 思路

　　因为不需要保证只有主对角线上有黑块

　　所以这是道蓝题

　　那么只要由让 S -> Line[i]，Row[j] ->T

　　在i == j 时给行列上连边即可

　　注意下因为不用保证只有主对角线上才有黑块

　　所以这样跑出来的Maxflow是有可能大于n的

　　注意一下输出条件即可

CODE

#include <bits/stdc++.h>
#define dbg(x) cout << #x << "=" << x << endl
#define eps 1e-8
#define pi acos(-1.0)

using namespace std;
typedef long long LL;

const int inf = 0x3f3f3f3f;

template<class T>inline void read(T &res)
{
    char c;T flag=1;
    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
    while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
}

namespace _buff {
    const size_t BUFF = 1 << 19;
    char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
    char getc() {
        if (ib == ie) {
            ib = ibuf;
            ie = ibuf + fread(ibuf, 1, BUFF, stdin);
        }
        return ib == ie ? -1 : *ib++;
    }
}

int qread() {
    using namespace _buff;
    int ret = 0;
    bool pos = true;
    char c = getc();
    for (; (c < '0' || c > '9') && c != '-'; c = getc()) {
        assert(~c);
    }
    if (c == '-') {
        pos = false;
        c = getc();
    }
    for (; c >= '0' && c <= '9'; c = getc()) {
        ret = (ret << 3) + (ret << 1) + (c ^ 48);
    }
    return pos ? ret : -ret;
}

const int maxn = 200007;

int n, m;
int s, t;

struct edge{
    int from,to;
    LL cap,flow;
};

map<int, int> vis;

struct DINIC {
    int head[maxn << 1], nxt[maxn << 1], edge[maxn << 1], cnt;
    int cap[maxn << 1], depth[maxn << 1];

    void init() {
        cnt = 1;
        memset(head, 0, sizeof(head));
    }

    void BuildGraph(int u, int v, int w) {
        ++cnt;
        edge[cnt] = v;
        nxt[cnt] = head[u];
        cap[cnt] = w;
        head[u] = cnt;

        ++cnt;
        edge[cnt] = u;
        nxt[cnt] = head[v];
        cap[cnt] = 0;
        head[v] = cnt;
    }

    queue<int> q;

    bool bfs() {
        memset(depth, 0, sizeof(depth));
        depth[s] = 1;
        q.push(s);
        while(!q.empty()) {
            int u = q.front();
            q.pop();
            for ( int i = head[u]; i; i = nxt[i] ) {
                int v = edge[i];
                if(depth[v]) {
                    continue;
                }
                if(cap[i]) {
                    depth[v] = depth[u] + 1;
                    q.push(v);
                }
            }
        }
        return depth[t];
    }

    int dfs(int u, int dist) {
        if(u == t) {
            return dist;
        }
        int flow = 0;
        for ( int i = head[u]; i && dist; i = nxt[i] ) {
            if(cap[i] == 0)
                continue;
            int v = edge[i];
            if(depth[v] != depth[u] + 1) {
                continue;
            }
            int res = dfs(v, min(cap[i], dist));
            cap[i] -= res;
            cap[i ^ 1] += res;
            //printf("cap[%d]:%d
",t, cap[t]);
            dist -= res;
            flow += res;
        }
        return flow;
    }

    int maxflow() {
        int ans = 0;
        while(bfs()) {
            ans += dfs(s, inf);
        }
        return ans;
    }
} dinic;

int main()
{
    //freopen("data.txt", "r", stdin);
    int T;
    read(T);
    while(T--) {
        read(n);
        dinic.init();
        s = 0, t = n * 2 + 1;
        int x;
        for ( int i = 1; i <= n; ++i ) {
            dinic.BuildGraph(s, i, 1);
            dinic.BuildGraph(n + i, t, 1);
            for ( int j = 1; j <= n; ++j ) {
                read(x);
                if(x == 1) {
                    dinic.BuildGraph(i, n + j, 1);
                }
            }
        }
        int ans = dinic.maxflow();
        //dbg(ans);
        if(ans < n) {
            puts("No");
        }
        else {
            puts("Yes");
        }
    }
    return 0;
}