ZOJ 1298 Domino Effect

最新推荐文章于 2015-12-26 17:06:00 发布

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本文介绍了一种利用Dijkstra算法求解最长路径的方法。通过先计算从起点1出发到达所有点的最短路径，再枚举每条边判断是否能形成更长的有效路径，并最终输出最长路径的一半作为结果。

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首先dijkstra求出最短路,求的时候顺便求出1能到的最长路记为ans * 2.

枚举每条边,满足dis[i] + dis[j] + g[i][j] > ans 更新ans.

最后输出答案ans / 2;

/*
 * ZOJ 1298.cpp
 *
 *  Created on: May 27, 2013
 *      Author: root
 */
#include <iostream>
#include <cstdio>
#include <queue>
#include <memory.h>
#include <limits.h>
using namespace std;

const int maxn = 505;
struct edge{
	int v, next;
	double w;
}es[maxn * 5];

struct node{
	int i;
	double v;
	node(int ii = 0, double vv = 0):i(ii), v(vv){}
	bool operator<(const node & rhs)const{
		return v > rhs.v;
	}
};

int head[maxn], n, m, fro, tar, type;
double ans;
double dis[maxn];
bool vis[maxn];

void addEdge(int u, int v, double w, int eidx){
	es[eidx].v = v;
	es[eidx].w = w;
	es[eidx].next = head[u];
	head[u] = eidx;
}
void dijkstra(){
	type = 0;
	for(int i = 1; i <= n; ++i){
		dis[i] = INT_MAX;
	}
	memset(vis, 0, sizeof(vis));
	dis[1] = 0;
	priority_queue<node> q;
	q.push(node(1, 0));
	while(q.size()){
		node t = q.top();
		q.pop();
		int u = t.i;
		if(vis[u])continue;
		vis[u] = 1;
		if(t.v >= ans){
			fro = u;
			ans = t.v;
		}
		for(int ne = head[u]; ne != -1; ne = es[ne].next){
			int v = es[ne].v;
			double w = es[ne].w;
			if(dis[v] > dis[u] + w && !vis[v]){
				dis[v] = dis[u] + w;
				q.push(node(v, dis[v]));
			}
		}
	}
}
int main(){
	int cas = 1;
	while(scanf("%d %d", &n, &m)){
		if(!n && !m)break;
		memset(head, -1, sizeof(head));
		int eidx = 0;
		for(int i = 0; i < m; ++i){
			int u, v;
			double w;
			scanf("%d %d %lf", &u, &v, &w);
			addEdge(u, v, w, eidx++);
			addEdge(v, u, w, eidx++);
		}
		ans = 0;
		fro = 1;
		dijkstra();

		ans *= 2;

		for(int i = 1; i <= n; ++i){
			for(int ne = head[i]; ne != -1; ne = es[ne].next){
				int u = i, v = es[ne].v;
				double w = es[ne].w;
				if(dis[u] + dis[v] + w > ans){
					fro = u, tar = v;
					type = 1;
					ans = dis[u] + dis[v] + w;
				}
			}
		}

		printf("System #%d\n", cas++);
		if(type == 0){
			if(m == 0) fro = 1;
			printf("The last domino falls after %.1lf seconds, at key domino %d.\n", ans / 2, fro);
		}else{
			printf("The last domino falls after %.1lf seconds, between key dominoes %d and %d.\n", ans / 2, fro, tar);
		}
		printf("\n");
	}
	return 0;
}