POJ 1062 昂贵的聘礼(dij+邻接矩阵)

( ̄▽ ̄)"

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;

const int INF=10e8;
const int MAXN=110;
int k,minn;
int c[MAXN][MAXN],lc[MAXN],level[MAXN],X[MAXN];
bool vis[MAXN];

int dij(int n)
{
    for(int i=1;i<=n;i++)
        lc[i]=c[0][i];
    for(int j=1;j<=n;j++)
    {
        k=-1,minn=INF;
        for(int i=1;i<=n;i++)
        {
            if(!vis[i]&&lc[i]<minn)
            {
                minn=lc[i];
                k=i;
            }
        }
        if(k==-1) break;
        vis[k]=1;
        for(int i=1;i<=n;i++)
            if(!vis[i]&&c[k][i]>0&&lc[i]>lc[k]+c[k][i]) //c[k][i]>0¶ø²»ÊÇc[k][i]>=0
                lc[i]=lc[k]+c[k][i];
    }
    return lc[1];
}

int main()
{
    int m,n;
    scanf("%d%d",&m,&n);
    memset(c,0,sizeof(c));
    memset(level,0,sizeof(level));
    memset(lc,INF,sizeof(lc));
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d%d",&c[0][i],&level[i],&X[i]);
        int t,v;
        for(int j=1;j<=X[i];j++)
        {
            scanf("%d%d",&t,&v);
            c[t][i]=v;
        }
    }
    int maxLevel,minPrice=INF;
    for(int i=1;i<=n;i++)
    {
        maxLevel=level[i];
        for(int j=1;j<=n;j++)
        {
            if(level[j]>maxLevel||maxLevel-level[j]>m)
                vis[j]=1;
            else vis[j]=0;
        }
        int price=dij(n);
        minPrice=min(minPrice,price);
    }
    printf("%d\n",minPrice);
    return 0;
}
posted @ 2016-02-17 23:50  &ATM  阅读(176)  评论(0编辑  收藏  举报
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