poj 1062 最短路（昂贵的聘礼）

来源：互联网发布：em算法应用实例编辑：程序博客网时间：2024/06/09 18:50

题意：中文题。

思路：最短路dijkstra。需要枚举可行区间。每次运行不用重复构图，只需将不符合题意的点在used数组提前赋值为1即可。

#include <stdio.h>#include <string.h>#define INF 0x3fffffff#define min(a,b) ((a)<(b)?(a):(b))#define max(a,b) ((a)>(b)?(a):(b))#define N 105int lev[N],g[N][N],dis[N],used[N];int n,m,k,diff;//diff为等级差别int dijkstra(int a,int b){    int i,j,min,res=0,now;    for(i = 0;i<=n;i++)        dis[i] = INF;    memset(used, 0, sizeof(used));    for(i = 1;i<=n;i++)        if(lev[i]<a || lev[i]>b)            used[i] = 1;    dis[0] = 0;    while(1){        min = INF;        for(j = 0;j<=n;j++)            if(!used[j] && dis[j]<min){                min = dis[j];                now = j;            }        if(min == INF)            break;        used[now] = 1;        res += min;        for(j = 1;j<=n;j++)            if(!used[j] && dis[j]>dis[now]+g[now][j])                dis[j] = dis[now]+g[now][j];    }    return dis[1];}int main(){    int i,j,a,b,res = INF;    scanf("%d %d",&diff,&n);    memset(lev, 0, sizeof(lev));    for(i = 0;i<=n;i++)        for(j = 0;j<=n;j++)            g[i][j] = INF;    for(i = 1;i<=n;i++){        scanf("%d %d %d",&a,&lev[i],&j);        g[0][i] = a;        while(j--){            scanf("%d %d",&a,&b);            g[a][i] = b;        }    }    for(i = max(0,lev[1]-diff);i<=lev[1];i++){        j = dijkstra(i, i+diff);        res = min(res,j);    }    printf("%d\n",res);    return 0;}

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