poj 1062 昂贵的聘礼

http://poj.org/problem?id=1062

题意：一个人想娶酋长的女儿，酋长有个条件，就是用金币或者是用物品和少量的金币来交换，其他拥有物品的人亦可以用同样方法来交换；

思路：dijkstra+枚举，思路是从网上看的，题目中的“地位较高的的人不会再和他交易”其实挺干扰人的，没有用到这个；

代码：

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <cstring>
using namespace std;
int leval[110] = {0};
int value[110] = {0};
int map[110][110] = {0};
int ok[110] = {0};
int n = 0;
int len = 0;
int  p = 0;
int val = 0;
int vis[110] = {0};
int d[110] = {0};
int dijkstra()
{
    memset(vis,0,sizeof(vis));
    d[1] = 0;
    for(int i = 2;i <= n; ++i)
    d[i] = 99999999;
    for(int i = 1; i <= n; ++i)
    {
        int temp = 0;
        int max = 99999999;
        for(int j = 1;j <= n; ++j)
        if(ok[j] && !vis[j] && d[j] < max)
        {
            max = d[j];
            temp = j;
        }
        vis[temp] = 1;
        for(int j = 1;j <= n; ++j)
        if(ok[j]&&!vis[j])
        d[j] = min(d[j],map[temp][j] + d[temp]);
    }
    int mm = d[1] + value[1];
    for(int i = 2;i <= n; ++i)
    {
       d[i] = d[i] + value[i];
       if(d[i] < mm)
       mm = d[i];
    }
    return mm;
}
int main()
{
    int num = 0;
    scanf("%d%d",&len,&n);
    for(int i = 0 ;i <= n; ++i)
    for(int j = 0; j <= n; ++j)
    map[i][j] = 99999999;
    for(int i = 0;i <= n; ++i)
    map[i][i] = 0;
    for(int i = 1;i <= n; ++i)
    {
        scanf("%d%d%d",&value[i],&leval[i],&num);
        for(int j = 1;j <= num; ++j)
        {
            scanf("%d%d",&p,&val);
            map[i][p] = val;
        }
    }
    int tt = leval[1];
    int mincost = 999999999;
    for(int i = 0;i <= len; ++i)
    {
        memset(ok,0,sizeof(ok));
            for(int j = 1;j <= n; ++j)
            if(leval[j] >= tt-len+i&&leval[j] <= tt+i)
            ok[j] = 1;
           int cost= dijkstra();
          if(cost < mincost)
          mincost = cost;
    }
    printf("%d\n",mincost);
    return 0;
}