Luogu P3254 圆桌问题

题目链接 \(Click\) \(Here\)

水题。记得记一下边的流量有没有跑完。

#include <bits/stdc++.h>
using namespace std;

const int N = 100010;
const int M = 800010;
const int INF = 0x3f3f3f3f;

int cnt = -1, head[N];

struct egde {
	int nxt, to, f;
}e[M];

void add_edge (int from, int to, int flw) {
	e[++cnt].nxt = head[from];
	e[cnt].to = to;
	e[cnt].f = flw;
	head[from] = cnt;
}

void add_len (int u, int v, int f) {
	add_edge (u, v, f);
	add_edge (v, u, 0);
}

int n, m, sum, id[210][310], hav[N], tak[N];

int nd1 (int x) {return x;}
int nd2 (int x) {return n + x;}

queue <int> q;
int cur[N], deep[N];

bool bfs (int s, int t) {
    memcpy (cur, head, sizeof (head));
    memset (deep, 0x3f, sizeof (deep));
    q.push (s); deep[s] = 0;
    while (!q.empty ()) {
        int u = q.front (); q.pop ();
        for (int i = head[u]; ~i; i = e[i].nxt) {
            int v = e[i].to;
            if (deep[v] == INF && e[i].f) {
                deep[v] = deep[u] + 1;
                q.push (v);
            }
        }
    }
    return deep[t] != INF;
}

int dfs (int u, int t, int lim) {
    if (u == t || !lim) {
        return lim;
    }
    int tmp = 0, flow = 0;
    for (int &i = cur[u]; ~i; i = e[i].nxt) {
        int v = e[i].to;
        if (deep[v] == deep[u] + 1) {
            tmp = dfs (v, t, min (lim, e[i].f));
            lim -= tmp;
            flow += tmp;
            e[i ^ 0].f -= tmp;
            e[i ^ 1].f += tmp;
            if (!lim) break;
        }
    }
    return flow;
}

int main () {
	memset (head, -1, sizeof (head));
	cin >> n >> m;
	for (int i = 1; i <= n; ++i) {
		cin >> hav[i];
		sum += hav[i];
		if (hav[i] > m) {
			puts ("0");
			return 0;
		}
	}
	for (int i = 1; i <= m; ++i) {
		cin >> tak[i];
	}
	int s = n + m + 1;
	int t = n + m + 2;
	for (int i = 1; i <= n; ++i) add_len (s, nd1 (i), hav[i]);
	for (int i = 1; i <= m; ++i) add_len (nd2 (i), t, tak[i]);
	
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			id[i][j] = cnt + 1;
			add_len (nd1 (i), nd2 (j), 1);
		}
	}
	int max_flow = 0;
	while (bfs (s, t)) {
		max_flow += dfs (s, t, INF);
	}
	if (max_flow != sum) {puts ("0"); return 0;}
	puts ("1");
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			if (!e[id[i][j]].f) {
				printf ("%d ", j);
			}
		}
		printf ("\n");
	}
} 

posted @ 2019-03-05 21:23  maomao9173  阅读(87)  评论(0编辑  收藏  举报