PKU 1062 昂贵的聘礼

最新推荐文章于 2019-10-23 15:45:10 发布

转载最新推荐文章于 2019-10-23 15:45:10 发布 · 80 阅读

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原文链接：http://www.cnblogs.com/lotus3x/archive/2008/07/23/1249469.html

/*
枚举等级范围, 每种情况求最短路
*/

#include <iostream>
#include <queue>
#define MAXN 101
using namespace std;

int MAP[MAXN][MAXN];
int D[MAXN];

int LEV;
int p[MAXN], l[MAXN];
int e[10000][3];

int bellman(int s, int t, int n)
{
    bool mk[MAXN] = {false};
    queue<int> q;
    int i, x, tmp;
    for (D[s] = 0, q.push(s); !q.empty(); )
        for (x = q.front(), q.pop(), mk[x] = false, i = 0; i < n; i++)
            if (MAP[x][i] != INT_MAX && (tmp = D[x] + MAP[x][i]) < D[i])
            {
                D[i] = tmp;
                if (!mk[i])
                    mk[i] = true, q.push(i);
            }
    return D[t];
}

int main()
{

    int n, i, j, m, w, cur = 0;
    int x, y;
    int king, lev;
    int cost, min_cost;

    scanf("%d %d", &LEV, &n);

    for (i = 0; i <= n; i++)
        MAP[i][0] = INT_MAX;

    for (i = 1; i <= n; i++)
    {
        scanf("%d %d %d", &p[i], &l[i], &m);
        while (m--)
        {
            scanf("%d %d", &j, &w);
            e[cur][0] = i;
            e[cur][1] = j;
            e[cur][2] = w;
            cur++;
        }
        MAP[i][0] = p[i];
    }
    king = l[1];
    min_cost = p[1];

    for (lev = king - LEV; lev <= king; lev++)
    {
        for (i = 0; i <= n; i++)
            for (D[i] = INT_MAX, j = 1; j <= n; j++)
                MAP[i][j] = INT_MAX;

        for (i = 0; i < cur; i++)
        {
            x = e[i][0];
            y = e[i][1];
            if (l[x] >= lev && l[x] <= lev+LEV
                && l[y] >= lev && l[y] <= lev+LEV)
                MAP[x][y] = e[i][2];
        }
        cost = spfa(1, 0, n+1);
        if (cost < min_cost)
            min_cost = cost;
    }
    printf("%d\n", min_cost);
    return 0;
}

转载于:https://www.cnblogs.com/lotus3x/archive/2008/07/23/1249469.html