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loj#10078. 新年好(最短路)

题目:

loj#10078. 新年好

解析:

亲戚只有五个,可以把它们看成2,3,4,5,6号点,分别跑最短路,记录一下距离,然后DFS一下
这题非常玄学,我开了一个\(12*12\)的数组,没有离散化,竟然过了,开到\(5050*5050\)就RE,玄学

代码:

#include <bits/stdc++.h>
using namespace std;

const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;

int n, m, num, cnt, ans = INF;
int head[N], dis[N], home[N], d[12][12];

bool vis[N], mark[N];

struct node {
	int v, nx, w;
} e[N];

struct edge {
	int id, dis;
	bool operator <(const edge &oth) const {
		return this -> dis > oth.dis;
	}
};

inline void add(int u, int v, int w) {
	e[++num] = (node) {v, head[u], w}, head[u] = num;
}

priority_queue<edge>q;
void dijkstra(int S) {
	memset(dis, INF, sizeof dis);
	memset(vis, 0, sizeof vis);
	dis[S] = 0;
	q.push((edge) {S, 0});
	while (!q.empty()) {
		edge d = q.top();
		q.pop();
		int u = d.id;
		if (vis[u]) continue;
		vis[u] = 1;
		for (int i = head[u]; ~i; i = e[i].nx) {
			int v = e[i].v;
			if (dis[v] > dis[u] + e[i].w)
				q.push((edge) {v, dis[v] = dis[u] + e[i].w});
		}
	}
}

void dfs(int u, int sum, int cnt) {
	if (cnt == 5) {
		ans = min(ans, sum);
		return;
	}
	if (sum > ans) return;
	for (int i = 2; i <= 6; ++i) {
		if (!mark[i]) {
			mark[i] = 1;
			dfs(i, sum + d[u][i], cnt + 1);
			mark[i] = 0;
		}
	}
}

int main() {
	memset(head, -1, sizeof head);
	cin >> n >> m;
	for (int i = 1; i <= 5; ++i) cin >> home[i];
	for (int i = 1, x, y, z; i <= m; ++i) 
		cin >> x >> y >> z, add(x, y, z), add(y, x, z);
	dijkstra(1);
	for (int i = 1; i <= 5; ++i)
		d[1][i + 1] = d[i + 1][1] = dis[home[i]];
	for (int i = 1; i <= 5; ++i) {
		dijkstra(home[i]);
		for (int j = i + 1; j <= 5; ++j) 
			d[i + 1][j + 1] = d[j + 1][i + 1] = dis[home[j]];
	}
	
	mark[1] = 1;
	dfs(1, 0, 0);
	cout << ans << endl;
}
posted @ 2019-08-31 09:51  Chrety  阅读(...)  评论(...编辑  收藏