学步园

嗯 poj上很少见的中文题目，不会做，参考了别人的思路，这里就不写了，

地址是http://www.cnblogs.com/lyy289065406/archive/2011/07/30/2121678.html

写的很详细了，code

#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <limits>
#include <vector>
#include <bitset>
#include <string>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <string.h>
#include <iostream>
#include <algorithm>
#define Si set<int>
#define LL long long
#define pb push_back
#define PS printf(" ")
#define Vi vector<int>
#define LN printf("\n")
#define lson l,m,rt << 1
#define rson m+1,r,rt<<1|1
#define SD(a) scanf("%d",&a)
#define PD(a) printf("%d",a)
#define SET(a,b) memset(a,b,sizeof(a))
#define FF(i,a) for(int i(0);i<(a);i++)
#define FD(i,a) for(int i(a);i>=(1);i--)
#define FOR(i,a,b) for(int i(a);i<=(b);i++)
#define FOD(i,a,b) for(int i(a);i>=(b);i--)
#define readf freopen("input.txt","r",stdin)
#define writef freopen("output.txt","w",stdout)
const int maxn = 101;
const int INF = 0x3fffffff;
const int dx[]={0,1,0,-1};
const int dy[]={1,0,-1,0};
const double pi = acos(-1.0);
using namespace std;
int lim,N, edge[maxn][maxn], lv[maxn], x, dis[maxn], t_edge[maxn][maxn];
bool vis[maxn];
queue<int> q;
void solve(int x){
    FOR(i,0,N) FOR(j,0,N){
        if(lv[i]<lv[1]+x || lv[i]>lv[1]+x+lim || lv[j]<lv[1]+x || lv[j]>lv[1]+x+lim)
            t_edge[i][j]=INF;
        else
            t_edge[i][j]=edge[i][j];
    }
}
int spfa(){
    SET(vis,false);
    FOR(i,1,N) dis[i]=INF;
    dis[0]=0;
    vis[0]=true;
    while(!q.empty()) q.pop();
    q.push(0);

    while(!q.empty()){
        int tmp=q.front();
        q.pop();
        vis[tmp]=false;

        FOR(i,1,N){
            if(t_edge[tmp][i]==INF) continue;

            if(dis[i] > dis[tmp]+ t_edge[tmp][i]){
                dis[i]=dis[tmp]+ t_edge[tmp][i];

                if(!vis[i]){
                    vis[i]=true;
                    q.push(i);
                }
            }

        }
    }

    return dis[1];
}
int main(){

    scanf("%d%d",&lim,&N);
    FOR(i,0,N) FOR(j,0,N) edge[i][j]=INF;
    FOR(i,1,N){
        scanf("%d%d%d",&edge[0][i],&lv[i],&x);

        FF(j,x){
            int t;
            SD(t);SD(edge[t][i]);
        }
    }
    lv[0]=lv[1];
    int minn=INF;
    FOR(i,-lim,0){
        solve(i);
        int tt=spfa();
        minn=min(tt,minn);
    }
    PD(minn);LN;
	return 0;
}