[poj 1062] 昂贵的聘礼

题目地址 ： http://poj.org/problem?id=1062

一开始理解错误了，以为只要是相邻节点的等级差不大于等级限制就ok，原来是必须这个方案中的节点等级差都必须不大于等级限制。

解题思路：因为必须包含1号节点（酋长），所以把所有关于1号节点范围内的等级都列举出来，每次都是不同的图，取最小的就ok了

用的邻接表 + spfa，上代码

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <queue>
#define N   102
#define INF 200000000

int m, n;

int p[N];
int level[N];
int cm[N];
bool alreadyInQueue[N];

int max_level;

int head[N];
int next[N*N];
int edge[N*N];
int cost[N*N];

void addEdge(int x, int y, int c){
    
    static int i = 0;
    
    edge[i] = y;
    cost[i] = c;
    next[i] = head[x];
    head[x] = i;
    
    i++;
}

void spfa(){

    for (int i = 1; i <= n; i++){
        cm[i] = INF;
        alreadyInQueue[i] = false;
    }
    
    cm[1] = 0;
    alreadyInQueue[1] = true;
    
    std::queue<int> q;
    q.push(1);
    
    while(!q.empty()){
        int x = q.front();
        q.pop();
        alreadyInQueue[x] = false;
        
        for (int e = head[x]; e != -1; e = next[e]){
            int y = edge[e];
            
            if ( level[y] <= max_level
               && max_level - level[y] <= m 
               && cm[x] + cost[e] < cm[y]){
                cm[y] = cm[x] + cost[e];
                
                if (!alreadyInQueue[y]){
                    q.push(y);
                    alreadyInQueue[y] = true;
                }
            }
        }    
    }
}


int main() {

    scanf("%d%d", &m, &n);
    memset(head, -1, sizeof(head));
    memset(next, -1, sizeof(next));
    
    for (int i = 1; i <= n; i++){
        int P, L, X;
        scanf("%d%d%d", &P, &L, &X);
        
        p[i] = P;
        level[i] = L;

        for (int j = 0; j < X; j++){
            int T, V;
            scanf("%d%d", &T, &V);
            addEdge(i, T, V);    
        }
    }
    
    int minimal = p[1];
    for (int i = level[1] + m; i >= level[1]; i--){
        max_level = i;
        
        spfa();
        
        for (int i = 2; i <= n; i++){
            if (cm[i] + p[i] < minimal){
                minimal = cm[i] + p[i];
            }
        }
    }

    printf("%d", minimal);
    
    return 0;
}