【Codeforces 283 C. Coin Troubles】+ 拓扑序 + 完全背包

C. Coin Troubles
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
In the Isle of Guernsey there are n different types of coins. For each i (1 ≤ i ≤ n), coin of type i is worth ai cents. It is possible that ai = aj for some i and j (i ≠ j).
Bessie has some set of these coins totaling t cents. She tells Jessie q pairs of integers. For each i (1 ≤ i ≤ q), the pair bi, ci tells Jessie that Bessie has a strictly greater number of coins of type bi than coins of type ci. It is known that all bi are distinct and all ci are distinct.
Help Jessie find the number of possible combinations of coins Bessie could have. Two combinations are considered different if there is some i (1 ≤ i ≤ n), such that the number of coins Bessie has of type i is different in the two combinations. Since the answer can be very large, output it modulo 1000000007 (109 + 7).
If there are no possible combinations of coins totaling t cents that satisfy Bessie’s conditions, output 0.
Input
The first line contains three space-separated integers, n, q and t (1 ≤ n ≤ 300; 0 ≤ q ≤ n; 1 ≤ t ≤ 105). The second line contains n space separated integers, a1, a2, …, an (1 ≤ ai ≤ 105). The next q lines each contain two distinct space-separated integers, bi and ci (1 ≤ bi, ci ≤ n; bi ≠ ci).
It’s guaranteed that all bi are distinct and all ci are distinct.
Output
A single integer, the number of valid coin combinations that Bessie could have, modulo 1000000007 (109 + 7).
Examples
Input
4 2 17
3 1 2 5
4 2
3 4
Output
3
Input
3 2 6
3 1 1
1 2
2 3
Output
0
Input
3 2 10
1 2 3
1 2
2 1
Output
0
Note
For the first sample, the following 3 combinations give a total of 17 cents and satisfy the given conditions: {0 of type 1, 1 of type 2, 3 of type 3, 2 of type 4}, {0, 0, 6, 1}, {2, 0, 3, 1}.
No other combinations exist. Note that even though 4 occurs in both bi and ci,  the problem conditions are still satisfied because all bi are distinct and all ci are distinct.

对于 u，v ,选 v 必选 u ，且 num[u] > num[v],所以 u 至少有一个，而 vl[v] += vl[u],选 v 必须选 u，有可能会有 u v,v u 的环，此时满足不了，而后完全背包

AC代码：

#include<cstdio>#include<cstring>using namespace std;const int mod = 1e9 + 7;const int K = 1e5 + 10;const int k = 310;int dp[K],in[k],s[k],a[k];int main(){    int n,q,t,u,v;    scanf("%d %d %d",&n,&q,&t);    for(int i = 1; i <= n; i++)        scanf("%d",&a[i]);        memset(in,0,sizeof(in));        memset(s,0,sizeof(s));        memset(dp,0,sizeof(dp));    for(int i = 1; i <= q; i++){        scanf("%d %d",&u,&v);        in[v]++,s[u] = v;    }    while(q--){        int p = 0;        for(int i = 1; i <= n; i++)            if(!in[i] && s[i]){                p = i; break;            }        if(!p) { puts("0"); return 0; } // 说明有环，不能满足        t -= a[p]; // 这枚硬币至少有一个        if(t < 0) { puts("0"); return 0; }        int ms = s[p];        s[p] = 0;        in[ms]--;        a[ms] += a[p]; // 不能比 p 少，选 ms 必选 p    }    dp[0] = 1;    for(int i = 1; i <= n; i++) // 完全背包        for(int j = a[i]; j <= t; j++)            dp[j] = (dp[j] + dp[j - a[i]]) % mod;    printf("%d\n",dp[t]);    return 0;}