poj 3255

题意：次短路

雾草.....当时卡了好久呢
题解：开两个数组，dist用来储存第一次到达某个点的距离，dist2用来储存第二次到达某个点的距离，然后当到达某个点的距离既大于dist[i]和dist2[i]时就抛弃它，另外注意这个图中允许重复到达某个点

比如

1 2 100

2 3 200

那么1-3的次短路就是1->2->1->2->3

代码如下：

#include <stdio.h>
#include <algorithm>
#include <queue>
#include <vector>
#include <iostream>
#include <string.h>
using namespace std;
#define maxn 5010
#define inf 1e9
struct node{
    int to,cost;
};
typedef pair<int,int>p;
priority_queue<p,vector<p>,greater<p> >q;
int dist[maxn],dist2[maxn],n,r;
vector<node>Map[maxn];

void slove(){
    p dian;
    dian.first=0,dian.second=0;
    q.push(dian);
    dist[0]=0;
    while(!q.empty()){
        dian=q.top();
        q.pop();
        //if(dist2[dian.first]<dian.second)continue;
        for(int i=0;i<Map[dian.first].size();i++){
            node DIAN=Map[dian.first][i];
            int d=dian.second+DIAN.cost;
            if(dist[DIAN.to]>d){
                swap(dist[DIAN.to],d);
                q.push(p(DIAN.to,dist[DIAN.to]));
            }
            if(dist2[DIAN.to]>d && dist[DIAN.to]<d){
                dist2[DIAN.to]=d;
                q.push(p(DIAN.to,dist2[DIAN.to]));
            }
        }
    }
    if(n==1)dist2[n-1]=0;
    printf("%d\n",dist2[n-1]);
}

int main(){
    scanf("%d%d",&n,&r);
    fill(dist,dist+n,inf);
    fill(dist2,dist2+n,inf);
    for(int i=1;i<=r;i++){
        node now;
        int from;
        scanf("%d%d%d",&from,&now.to,&now.cost);
        from--;
        now.to--;
        Map[from].push_back(now);
        swap(from,now.to);
        Map[from].push_back(now);
    }
    slove();
}