POJ-2387-Til the Cows Come Home [最短路]

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题目传送门

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题意:输入一个无向图，求N到1的最短路径。

思路:最短路模板题。

#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <cmath>#include <queue>#include <map>#include <set>using namespace std;int mp[1200][1200], N, T;const int inf = 0x3f3f3f3f;void init(){    for (int i = 0; i <= 1000; i++)        for (int j = 0; j <= 1000; j++)            mp[i][j] = inf;}void dij(int x){    int dis[1200], book[1200];    memset(book,0,sizeof(book));    for (int i = 1; i <= N; i++)        dis[i] = mp[x][i];    book[x] = 1;    for (int i = 1; i <= N; i++)    {        int mi = inf, f = -1;        for (int j = 1; j <= N; j++)        {            if (!book[j] && dis[j]<mi)            {                mi = dis[j];                f = j;            }        }        if (f==-1)            break;        book[f]=1;        for (int j = 1; j <= N; j++)        {            if (!book[j] && dis[j]>dis[f]+mp[f][j])            {                dis[j] = dis[f]+mp[f][j];            }        }    }    printf("%d\n", dis[N]);    return;}int main(void){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    while (~scanf("%d %d", &T, &N))    {        init();        while (T--)        {            int x, y, s;            scanf("%d %d %d", &x, &y, &s);            if (mp[x][y]>s && mp[y][x]>s)                mp[x][y] = mp[y][x] = s;        }        dij(1);    }    return 0;   }