poj3255 Roadblocks

Posted 2020-08-29 多一份不为什么的坚持

Roadblocks Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13594   Accepted: 4783 Description Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather than the shortest path. She knows there must be some second-shortest path. The countryside consists of R (1 ≤ R ≤ 100,000) bidirectional roads, each linking two of the N (1 ≤ N ≤ 5000) intersections, conveniently numbered 1..N. Bessie starts at intersection 1, and her friend (the destination) is at intersection N. The second-shortest path may share roads with any of the shortest paths, and it may backtrack i.e., use the same road or intersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any other path). Input Line 1: Two space-separated integers: N and R  Lines 2..R+1: Each line contains three space-separated integers: A, B, and D that describe a road that connects intersections A and B and has length D (1 ≤ D ≤ 5000) Output Line 1: The length of the second shortest path between node 1 and node N Sample Input 4 4 1 2 100 2 4 200 2 3 250 3 4 100 Sample Output 450 Hint Two routes: 1 -> 2 -> 4 (length 100+200=300) and 1 -> 2 -> 3 -> 4 (length 100+250+100=450) Source USACO 2006 November Gold

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#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <map>
#define for(i,x,n) for(int i=x;i<n;i++)
#define ll long long int
#define INF 0x3f3f3f3f
#define MOD 1000000007

using namespace std;

int MAX_V,MAX_E;
struct edge{int to,cost;};
typedef pair<int,int> P;//最短距离，编号
vector<edge> G[5005];
int d[5005];//最短距离
int d2[5005];//次短距离

void dijkstra(int s){
    priority_queue<P,vector<P>,greater<P> > que;
    fill(d,d+MAX_V,INF);
    fill(d2,d2+MAX_V,INF);
    d[s]=0;
    que.push(P(d[s],s));
    while(!que.empty()){
        P p=que.top(); que.pop();
        int v=p.second;
        for(i,0,G[v].size()){
            int to=G[v][i].to;
            int dd=p.first+G[v][i].cost;
            if(dd<d[to]){
                d[to]^=dd;
                dd^=d[to];
                d[to]^=dd;
                que.push(P(d[to],to));
            }
            if(dd<d2[to] && dd>d[to]){
                d2[to]=dd;
                que.push(P(d2[to],to));
            }
        }
    }
}

int main()
{
    int x,y,z;
    scanf("%d %d",&MAX_V,&MAX_E);
    for(i,0,MAX_E){
        scanf("%d %d %d",&x,&y,&z);
        x-=1;
        y-=1;
        edge ee;
        ee.to=y; ee.cost=z;
        G[x].push_back(ee);
        ee.to=x; ee.cost=z;
        G[y].push_back(ee);
    }
    dijkstra(0);
    printf("%d",d2[MAX_V-1]);
    return 0;
}