I 題目描述

輸入描述:

The input consists of several test cases terminated by end-of-file.
 
The first line of each test case contains two integers n and m.
The second line contains n integers d_1, d_2, \dots, d_nd1​,d2​,…,dn​.
The i-th of the following m lines contains two integers a_iai​ and b_ibi​.

輸出描述:

For each test case, print "`Yes`" without quotes if it is possible. 
Otherwise, print "`No`" without quotes.

輸入

2 1
1 1
1 2
2 1
2 2
1 2
3 2
1 1 2
1 3
2 3

輸出

Yes
No
Yes

#include<bits/stdc++.h>
using namespace std;
const int inf = 1 << 30, N = 5e4 + 7, M = 3e5 + 7;
int head[N], ver[M], edge[M], nex[M], d[N], cur[N];
int n, m, s, t, tot, maxflow;
queue<int> q;
void init() {
    maxflow 
 
= 0;
    tot = 1;
    memset(head, 0, sizeof(head));
}
void add(int x, int y, int z) {
    ver[++tot] = y, edge[tot] = z, nex[tot] = head[x], head[x] = tot;
    ver[++tot] = x, edge[tot] = 0, nex[tot] = head[y], head[y] = tot;
}
bool bfs() {
    memset(d, 0, sizeof(d));
    while (q.size()) q.pop();
    q.push(s); d[s] = 1; cur[s] = head[s];
    while (q.size()) {
        int x = q.front(); q.pop();
        for (int i = head[x]; i; i = nex[i]) {
            int y = ver[i];
            if (edge[i] && !d[y]) {
                q.push(y);
                cur[y] = head[y];
                d[y] = d[x] + 1;
                if (y == t) return true;
            }
        }
    }
    return false;
}
int dinic(int x, int flow) {
    if (x == t) return flow;
    int rest = flow, k, i;
    for (i = cur[x]; i && rest; i = nex[i]) {
        int y = ver[i];
        if (edge[i] && d[y] == d[x] + 1) {
            k = dinic(y, min(rest, edge[i]));
            if (!k) d[y] = 0;
            edge[i] -= k;
            edge[i ^ 1] += k;
            rest -= k;
        }
    }
    cur[x] = i;
    return flow - rest;
}
int main() {
    while (scanf("%d%d", &n, &m) != EOF) {
        int p, sum = 0, x, y; s = 2 * n + 1; t = 2 * n + 2;
        init();
        for (int i = 1; i <= n; i++) {
            scanf("%d", &p);
            add(s, i, p);
            add(i + n, t, p);
            sum += p;
        }
        for (int i = 1; i <= m; i++) {
            scanf("%d%d", &x, &y);
            add(x, y + n, 1);
            add(y, x + n, 1);
        }
        int flow;
        while (bfs()) {
            while ((flow = dinic(s, inf)))
                maxflow += flow;
        }
        if (maxflow == sum)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}