无向图上的动态规划

牛客的题

 

 

 

 

 

有一种情况，就是k小于n时候，1到不了k，那就把dp全设置成负无穷（-1不行）这样就用不到了

 

 

 

#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
#include<cstdio>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 7;
const ll INF = 1e17 + 7;

ll dp[maxn];
ll map[220][220];
struct Node {
	int t, p, val;
}que[maxn];
bool bml(Node a, Node b) {
	return a.t < b.t;
}

int main() {
	int n, m, q;
	scanf("%d%d", &n, &m);
	for (int i = 0; i <= n; i++) {
		for (int j = 0; j <= n; j++) {
			if (i == j) map[i][j] = 0;
			else map[i][j] = INF;
		}
	}
	int x, y;
	for (int i = 0; i < m; i++) {
		scanf("%d %d", &x, &y);
		map[x][y] = map[y][x] = 1;
	}
	for (int k = 1; k <= n; k++) {
		for(int i=1;i<=n;i++){
			for (int j = 1; j <= n; j ++ ) {
				map[i][j] = min(map[i][j], map[i][k] + map[k][j]);
			}
		}
	}
	scanf("%d", &q);
	for (int i = 1; i <= q; i++) {
		scanf("%d%d%d", &que[i].t, &que[i].p, &que[i].val);
	}
	sort(que + 1, que + 1 + q, bml);


	for (int i = 0; i <= q; i++) dp[i] = -INF;

	que[0].p = 1;
	dp[0] = 0;
	ll ans = 0;
	for (int i = 1; i <= q; i++) {
		if (i > n) dp[i] = max(dp[i], dp[i - n] + que[i].val);

		for (int j = 1; j <= n && i - j >= 0; j++) {
			if (que[i - j].t + map[que[i].p][que[i - j].p] <= que[i].t) {
				dp[i] = max(dp[i], dp[i - j] + que[i].val);
			}
		}
		ans = max(ans, dp[i]);
	}
	printf("%lld\n", ans);
	return 0;
}