poj 1062 最短路（昂贵的聘礼）

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 105
int 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;
}