POJ

来源：互联网发布：财经日历软件编辑：程序博客网时间：2024/05/18 02:13

分析：

因为有等级限制，所以枚举等级范围，再用最短路判断最小金币即可

可以根据酋长的等级来缩小枚举的范围

#include <cstring>#include <cstdio>#include <queue>#include <iostream>#include <algorithm>using namespace std;#define INF 1e9#define maxn 50001int n , m ;struct node{    int dec;    int nxt;    int val;};int cnt;node data[maxn];int state[maxn];int head[maxn];int d[maxn];int vis[maxn];void addege(int u,int v,int w){    data[cnt].dec = v;    data[cnt].val = w;    data[cnt].nxt = head[u];    head[u] = cnt++;}void spfa(int x){    memset(vis,0,sizeof(vis));    for(int i = 0 ; i < maxn ; ++i)d[i] = INF;    queue<int>Q;    Q.push(0);    vis[0] = 1;    d[0] = 0;    while(!Q.empty())    {        int u =  Q.front();        Q.pop();        vis[u] = 0;        for(int i = head[u] ; ~i ; i = data[i].nxt)        {            int v = data[i].dec;            if( state[v] > n + x || state[v] < x )continue;            if( d[v] > d[u] + data[i].val )            {                d[v] = d[u] + data[i].val;                if(!vis[v])                {                    vis[v] = 1;                    Q.push(v);                }            }        }    }}int main(){        cin >> n >> m;        cnt = 0;        for(int i = 0 ; i < maxn ; ++i)head[i] = -1;        for(int i = 1 ; i <= m ; ++i)        {            int a,b,c;            scanf("%d %d %d",&a,&b,&c);            state[i] = b;            addege( 0 , i , a );            while( c-- )            {                int x,y;                scanf("%d %d",&x,&y);                addege( x , i , y );            }        }        int ans = INF ;        for(int i = state[1] - n ; i <= state[1] ; ++i){            spfa(i);            ans = min(ans,d[1]);        }        cout << ans << endl;}

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