HDU 1317 - XYZZY

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HDU 1317 - XYZZY

题意

有N个房间,每个房间都有能量,有正有负,现在要求能否到达第n个房间,并且中途能量都为正

思路

如果存在正环,就有无限能量,这时候能到就行。用SPFA判环,如果某个点入队次数超过n,说明有正环存在。那么从起点到这个点的权就是无限大。如果最后能更新n的权,说明能到n。

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <queue>
#define MP(a, b) make_pair(a, b)
using namespace std;
const int MAXN = 110 + 5;
const int INF = 0x3f3f3f3f;
 
int mp[MAXN][MAXN], n, d[MAXN], w[MAXN];
 
bool SPFA()
{
    queue<int> qu;
    int vis[MAXN], cnt[MAXN];
    memset(vis, 0, sizeof vis);
    memset(cnt, 0, sizeof cnt);
    d[1] = 100;
    qu.push(1);
    vis[1] = 1;
    while (!qu.empty())
    {
        int t = qu.front();
        qu.pop();
        vis[t] = 0;
        cnt[t]++;
        if (cnt[t] > n)
            continue;
        if (cnt[t] == n)
            d[t] = INF;
        for (int i = 1; i <= n; i++)
        {
            if (mp[t][i] && d[i] < d[t] + w[i] && d[t] + w[i] > 0)
            {
                d[i] = d[t] + w[i];
                if (i == n)
                    return true;
                if (!vis[i])
                {
                    qu.push(i);
                    vis[i] = 1;
                }
            }
        }
    }
    return false;
}
 
int main()
{
    //freopen("input.txt", "r", stdin);
    int i, j, a, b, wei, nn;
    while (scanf("%d", &n), n != -1)
    {
        memset(d, -INF, sizeof d);
        memset(mp, 0, sizeof mp);
        for (i = 1; i <= n; i++)
        {
            scanf("%d%d", &w[i], &nn);
            for (j = 0; j < nn; j++)
            {
                scanf("%d", &a);
                mp[i][a] = 1;
            }
        }
        printf("%s\n", SPFA() ? "winnable" : "hopeless");
    }
    return 0;
}

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