POJ 2907

来源：互联网发布：奥卡姆剃刀知乎编辑：程序博客网时间：2024/06/06 18:35

题意：平面上给你n个点。1为起始点，问你从1号点出发经过其他所有点再回到1号点的最短距离，两点之间的距离是曼哈顿距离（abs（x1-x2）+abs(y1-y2)）。

思路：DFS，算出每次的距离，取最短即可。

#include<cmath>#include<cstdio>#include<string>#include<cstring>#include<algorithm>#define INF 0x7fffffffusing namespace std;typedef struct POS{    int x, y;    int dist(const POS &A){        return abs(x-A.x) + abs(y-A.y);     }}Node;Node node[11], start;int vis[11], ans, N;void dfs(int dep, int len, Node tmp){    if(dep == N){        ans = min(ans, len+tmp.dist(start));        return ;    }    for(int i = 0;i < N;i ++){        if(!vis[i]){            vis[i] = 1;            dfs(dep+1, len+tmp.dist(node[i]), node[i]);            vis[i] = 0;        }    }}int main(){    int t, n, m;    /* freopen("in.c", "r", stdin); */    scanf("%d", &t);    while(t--){        scanf("%d%d", &n, &m);        scanf("%d%d", &start.x, &start.y);        scanf("%d", &N);        for(int i = 0;i < N;i ++) scanf("%d%d", &node[i].x, &node[i].y);        ans = INF;        dfs(0, 0, start);        printf("The shortest path has length %d\n", ans);    }    return 0;}

0 0