POJ.2387 Til the Cows Come Home (SPFA)

POJ.2387 Til the Cows Come Home (SPFA)

题意分析

1. 首先给出T和N，T代表边的数量，N代表图中点的数量
2. 图中边是双向边，并不清楚是否有重边，我按有重边写的。
3. 直接跑spfa,dij，floyd都可以。
4. 求1到N的最短路。

代码总览

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#define nmax 1005
#define inf 1e8+7
using namespace std;

int mp[nmax][nmax];
int n,m;
int shortdis[nmax];
void spfa(int s)
{
    bool isinqueue[nmax];
    queue<int> q;
    while(!q.empty()) q.pop();
    int now = s;
    memset(isinqueue,0,sizeof(isinqueue));
    for(int i = 1;i<=n;++i) shortdis[i] = inf;
    shortdis[s] = 0;
    q.push(s);
    isinqueue[s] = true;
    while(!q.empty()){
        now = q.front(); q.pop();
        isinqueue[now] = false;
        for(int i = 1;i<=n;++i){
            if(i != now && mp[now][i] + shortdis[now] < shortdis[i]){
                q.push(i);
                isinqueue[i] = true;
                shortdis[i] = shortdis[now] + mp[now][i];
            }
        }
    }
}
int main()
{
    while(scanf("%d %d",&m,&n) != EOF){
        for(int i = 1 ;i<=n;++i)
            for(int j = 1;j<=n;++j)
                mp[i][j] = inf;
        for(int i = 1;i<=m;++i){
            int x,y,dis;
            scanf("%d %d %d",&x,&y,&dis);
            int temp = mp[x][y];
            if(dis<temp){
                mp[x][y] = mp[y][x] = dis;
            }
        }
        spfa(1);
        printf("%d\n",shortdis[n]);
    }
    return 0;
}