POJ--1122--FDNY to the Rescue!【最短路】

来源：互联网发布：大数据薪资待遇编辑：程序博客网时间：2024/06/11 14:22

题意：给你一个邻接矩阵信息，某点发生火灾，告诉你一些位置有消防队，问各个消防队到火灾地点的最短时间，并输出最短路的路径，输出按最短时间由小到大排序。

就是一个最短路问题，输出路径，直接dijkstra了，1A还是挺爽的

#include<cstring>#include<string>#include<fstream>#include<iostream>#include<iomanip>#include<cstdio>#include<cctype>#include<algorithm>#include<queue>#include<map>#include<set>#include<vector>#include<stack>#include<ctime>#include<cstdlib>#include<functional>#include<cmath>using namespace std;#define PI acos(-1.0)#define MAXN 100010#define eps 1e-7#define INF 0x7FFFFFFF#define seed 131#define long long ll;#define unsigned long long ull;#define lson l,m,rt<<1#define rson r,m+1,rt<<1|1struct node{    int id, s, time, pathl;    int path[25];}ans[25];int edge[25][25];int vis[25];int dist[25];int path[25];int n;bool cmp(node x,node y){    return x.time<y.time;}void dijkstra(int s,int e){    int i, j;    for(i=1;i<=n;i++){        dist[i] = edge[s][i];    }    memset(vis,0,sizeof(vis));    memset(path,0,sizeof(path));    vis[s] = 1;    for(i=0;i<n-1;i++){        int temp = INF, k = -1;        for(j=1;j<=n;j++){            if(!vis[j]&&dist[j]<temp){                k = j;                temp = dist[j];            }        }        if(k==-1)   break;        vis[k] = 1;        for(j=1;j<=n;j++){            if(!vis[j]&&edge[k][j]!=INF&&dist[j]>dist[k]+edge[k][j]){                dist[j] = dist[k] + edge[k][j];                path[j] = k;            }        }    }}int main(){    int i, j, s, e, x;    scanf("%d",&n);    for(i=1;i<=n;i++){        for(j=1;j<=n;j++){            scanf("%d",&x);            if(x==-1)   edge[i][j] = INF;            else    edge[i][j] = x;        }    }    scanf("%d",&e);    int k = 0;    printf("Org\tDest\tTime\tPath\n");    while(scanf("%d",&s)!=EOF){        dijkstra(s,e);        ans[k].id = k + 1;        ans[k].s = s;        ans[k].time = dist[e];        ans[k].pathl = 1;        int ll = 1;        ans[k].path[0] = e;        int ee = e;        while(path[ee]!=0){            ans[k].path[ll++] = path[ee];            ee = path[ee];            ans[k].pathl++;        }        ans[k].path[ll++] = s;        ans[k].pathl++;        if(ans[k].pathl==2&&ans[k].path[0]==ans[k].path[1]) ans[k].pathl = 1;        k++;    }    sort(ans,ans+k,cmp);    for(i=0;i<k;i++){        printf("%d\t%d\t%d\t",ans[i].s,e,ans[i].time);        for(j=ans[i].pathl-1;j>0;j--){            printf("%d\t",ans[i].path[j]);        }        printf("%d\n",ans[i].path[j]);    }    printf("\n");    return 0;}/*60  3  4 -1 -1 -1-1 0  4  5 -1 -12  3  0 -1 -1  28  9  5  0  1 -17  2  1 -1  0 -15 -1  4  5  4  05 5*/

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