zoj 2792 && poj 2472 106 miles to Chicago

题意：最短路径的变种。

思路：任意最短路径算法均可解之，将权值累加求最小值修改为权值累乘求最大值即可。

代码：

#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>

using namespace std;

const int inf = 10000000;
const int maxn = 105;

struct node{
	int v;
	double w;
	
	node(int _v, double _w){
		v = _v;
		w = _w;
	}
};


int n,m;
vector<node> list[maxn];
double dist[maxn];
bool inq[maxn];


bool input(){
	cin>>n;
	if(n == 0) return false;
	cin>>m;	
	
	for(int i = 1; i <= n; i++) list[i].clear();
	int u,v;
	int w;
	for(int i = 0; i < m; i++){
		cin>>u>>v>>w;
		list[u].push_back(node(v,w/100.0));
		list[v].push_back(node(u,w/100.0));
	}
	return true;
}

void spfa(){
	queue<int> q;
	for(int i = 1; i <= n; i++){
		dist[i] = -inf; inq[i] = false;
	}
	dist[1] = 1;
	q.push(1);
	inq[1] = true;
	while(!q.empty()){
		int u = q.front(); q.pop(); inq[u] = false;
		
		for(int i = 0; i < list[u].size(); i++){
			int v = list[u][i].v;
			double w = list[u][i].w;
			
			if(dist[u] * w > dist[v]){
				dist[v] = dist[u] * w;
				if(!inq[v]){
					q.push(v); inq[v] = true;
				}
			}
		}
	}
	
}


void solve(){
	spfa();
	printf("%.6lf percent\n",dist[n]*100.0);
} 

int main(){
	while(input()){
		solve();
	}
	
	return 0;
}