BZOJ 2763: [JLOI2011]飞行路线 spfa dp

题目链接:

http://www.lydsy.com/JudgeOnline/problem.php?id=2763

题解:

d[x][kk]表示从s到x用了kk次免费机会的最少花费。

代码:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
#include<cstring>
#define mp make_pair
#define X first
#define Y second
using namespace std;

const int maxn = 10000;

int n, m, k;
vector<pair<int, int> > G[maxn];

int d[maxn][22];
bool inq[maxn][22];
void spfa(int s) {
	memset(d, 0x7f, sizeof(d));
	memset(inq, 0, sizeof(inq));
	queue<pair<int,int> > Q;
	d[s][0] = 0, inq[s][0] = 1, Q.push(mp(s,0));
	while (!Q.empty()) {
		int u = Q.front().X,kk=Q.front().Y;
		Q.pop();
		inq[u][kk] = 0;
		for (int i = 0; i < G[u].size(); i++) {
			int v = G[u][i].X, w = G[u][i].Y;
			if (d[v][kk]>d[u][kk] + w) {
				d[v][kk] = d[u][kk] + w;
				if (!inq[v][kk]) {
					inq[v][kk] = 1, Q.push(mp(v, kk));
				}
			}
			if (kk + 1 <= k&&d[v][kk + 1] > d[u][kk]) {
				d[v][kk + 1] = d[u][kk];
				if (!inq[v][kk + 1]) {
					inq[v][kk + 1] = 1, Q.push(mp(v, kk + 1));
				}
			}
		}
	}
}

void init() {
	for (int i = 0; i < n; i++) G[i].clear();
}

int main() {
	while (scanf("%d%d%d", &n, &m, &k) == 3) {
		init();
		int s, t;
		scanf("%d%d", &s, &t);
		for (int i = 0; i < m; i++) {
			int u, v, w;
			scanf("%d%d%d", &u, &v, &w);
			G[u].push_back(mp(v, w));
			G[v].push_back(mp(u, w));
		}
		spfa(s);
		printf("%d\n", d[t][k]);
	}
	return 0;
}

  

 

posted @ 2016-06-23 00:51  fenicnn  阅读(123)  评论(0编辑  收藏  举报