Learning4Fun ural 1277

/*题意：有n个城市，m条边。现在有一群坏人从s到f城市，如果小防止坏人从i城市经过需要在该城市
 * 拍前ki个警察。现在总共有k个警察问能否防止坏人从s到达f
 * 解法：把n个城市拆成两个点建立边容量为Ki,其他边容量inf。看s到f的最大流是否小与k
 */
#include <functional>
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <math.h>
#include <ctime>

#include <list>
#include <set>
#include <map>
#include<time.h>
using namespace std;
#define N 250
#define E 105*105*16

#define typec int                   // type of cost
const typec inf = 0x3f3f3f3f; // max of cost
struct edge
{
        int x, y, nxt;
        typec c;
} bf[E];
int ne, head[N], cur[N], ps[N], dep[N];
void addedge(int x, int y, typec c)
{ // add an arc(x -> y, c);vertex: 0 ~ n-1;
    bf[ne].x = x;
    bf[ne].y = y;
    bf[ne].c = c;
    bf[ne].nxt = head[x];
    head[x] = ne++;
    bf[ne].x = y;
    bf[ne].y = x;
    bf[ne].c = 0;
    bf[ne].nxt = head[y];
    head[y] = ne++;
}
typec flow(int n, int s, int t)
{
    typec tr, res = 0;
    int i, j, k, f, r, top;
    while (1)
    {
        memset(dep, -1, n * sizeof(int));
        for (f = dep[ps[0] = s] = 0, r = 1; f != r;)
            for (i = ps[f++], j = head[i]; j; j = bf[j].nxt)
            {
                if (bf[j].c && -1 == dep[k = bf[j].y])
                {
                    dep[k] = dep[i] + 1;
                    ps[r++] = k;
                    if (k == t)
                    {
                        f = r;
                        break;
                    }
                }
            }
        if (-1 == dep[t])
            break;
        memcpy(cur, head, n * sizeof(int));
        for (i = s, top = 0;;)
        {
            if (i == t)
            {
                for (k = 0, tr = inf; k < top; ++k)
                    if (bf[ps[k]].c < tr)
                        tr = bf[ps[f = k]].c;
                for (k = 0; k < top; ++k)
                    bf[ps[k]].c -= tr, bf[ps[k] ^ 1].c += tr;
                res += tr;
                i = bf[ps[top = f]].x;
            }
            for (j = cur[i]; cur[i]; j = cur[i] = bf[cur[i]].nxt)
                if (bf[j].c && dep[i] + 1 == dep[bf[j].y])
                    break;
            if (cur[i])
            {
                ps[top++] = cur[i];
                i = bf[cur[i]].y;
            }
            else
            {
                if (0 == top)
                    break;
                dep[i] = -1;
                i = bf[ps[--top]].x;
            }
        }
    }
    return res;
}
int base[N];
int main()
{
    //freopen("data.in", "r", stdin);
    int k, n,m,s, f;
    ne = 2;
    memset(head, 0, sizeof(head));
    cin >> k;
    cin >> n >> m >> s >> f;

    for (int i = 1; i <= n; i++)
    {
        cin >> base[i];
        if(i==s||i==f)base[i]=inf;
        addedge(2 * i - 1, 2 * i, base[i]);

    }
    for (int i = 0; i < m; i++)
    {
        int x, y;
        cin >> x >> y;
        addedge(2 * x, 2 * y - 1, inf);
        addedge(2 * y, 2 * x - 1, inf);
    }
    n = 2 * n;
    int res=flow(n+1,2 * s - 1, 2 * f);
    if(res>k)cout<<"NO"<<endl;
    else cout<<"YES"<<endl;

}