POJ 1062 昂贵的聘礼(最短路径Dijkstra+枚举)

//最短路径——Dijkstra算法
//此题的关键在于等级限制的处理，最好的办法是采用枚举，即假设酋长等级为5，等级限制为2，那么需要枚举等级从3~5,4~6,5~7
//从满足改等级范围的结点组成的子图中用Dijkstra来算出最短路径
//小结，通过枚举的方式可以消除一些图与图之间的限制
#include<iostream>
#include<cmath>
#define INF 200000000
#define MAX 101
using namespace std;
int map[MAX][MAX],lev[MAX],d[MAX],value[MAX];
bool within_lim[MAX],v[MAX];//within_lim为满足等级限制的标记数组
int lev_lim,n;
int dijkstra()//Dijkstra算法
{
int minimum = INF;
memset(v,0,sizeof(v));//清除所有点的标号
for(int i = 1;i <= n;++i)
d[i] = (i == 1 ? 0 : INF);//设d[0] = 0,其他d[i] = INF
for(int i = 1;i <= n;++i)//循环N次
{
int x = 0, m = INF;
for(int y = 1; y <= n;++y)
if(!v[y] && d[y] <= m && within_lim[y])//在所有未标号且满足等级限制的结点中，选出d值最小的结点x
{
x = y;
m = d[y];
}
v[x] = 1;//给结点x标记
for(int y = 1;y <= n;++y)//对于从x出发的所有边(x,y)，更新d[y] = min{d[y], d[x] + map[x][y])
{
if(within_lim[y])//满足等级限制
d[y] = min(d[y],d[x] + map[x][y]);//更新d[y]值
}
}
for(int i = 1;i <= n;++i)
{
d[i] += value[i];//对于每个d[i]值，还需加上进入该结点的花费，再进行比较
if(d[i] < minimum)	minimum = d[i];
}
return minimum;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
cin >> lev_lim >> n;
for(int i = 0;i <= n;++i)
for(int j = 0;j <= n;++j)
map[i][j] = (i == j ? 0 : INF);//图的初始化，注意对角线初始化为0，从自己出发到自己的花费为0
for(int i = 1;i <= n;++i)
{
int t;
cin >> value[i] >> lev[i] >> t;
for(int j = 1;j <= t;++j)
{
int k;
cin >> k;
cin >> map[i][k];
}
}//建图完毕

int kinglev = lev[1];
int min_cost = INF,cost;
for(int i = 0;i <= lev_lim;++i)
{
memset(within_lim,0,sizeof(within_lim));//初始化标记数组
for(int j = 1;j <= n;++j)
if(lev[j] >= kinglev - lev_lim + i && lev[j] <= kinglev + i)//枚举等级允许范围的结点
within_lim[j] = 1;

cost = dijkstra();
if(cost < min_cost)
min_cost = cost;
}
cout << min_cost << endl;
return 0;
}