POJ-1062 昂贵的聘礼 区间枚举

第二次来做这一题,由于题目中给定了一个等级限制M,所以可以通过枚举第一个点所在的位置求解.思路很清晰.

for (int i = 0; i <= M; ++i) 这个i来表示第一个点的等级在M长度区间内的偏移量.然后再在区间内建边,floyd即可.

代码如下:

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#define INF 0x3f3f3f3f
using namespace std;

int M, N, mey[105], lvl[105], head[105], idx;

int lwall, rwall, G[105][105];

struct Edge {
    int v, val, next;
}e[10005]; // 保留替换列表 

void addedge(int x, int v, int val) {
    ++idx;
    e[idx].v = v, e[idx].val = val;
    e[idx].next = head[x], head[x] = idx;
}

inline bool in(int x) {
    if (lvl[x] >= lwall && lvl[x] <= rwall) {
        return true;    
    } else return false;
}

void build() {
    memset(G, 0x3f, sizeof (G));
    for (int i = 1; i <= N; ++i) {
        G[i][i] = 0;
        for (int j = head[i]; j != -1; j = e[j].next) {
            if (in(i) && in(e[j].v)) {
                G[i][e[j].v] = e[j].val;
            }
        }    
    }
}

void floyd(int &ret) {
    for (int k = 1; k <= N; ++k) {
        for (int i = 1; i <= N; ++i) {
            for (int j = 1; j <= N; ++j) {
                if (G[i][k]!=-1&&G[k][j]!=-1) { // 有边相连 
                    G[i][j] = min(G[i][j], G[i][k]+G[k][j]);
                }
            }    
        }    
    }
    for (int i = 1; i <= N; ++i) {
        if (G[1][i] != INF) {
            ret = min(ret, G[1][i]+mey[i]);
        }
    }
}

int main() {
    int c, v, val, ret;
    while (scanf("%d %d", &M, &N) == 2) {
        ret = INF;
        idx = -1;
        memset(head, 0xff, sizeof (head));
        for (int i = 1; i <= N; ++i) {
            scanf("%d %d %d", &mey[i], &lvl[i], &c);
            for (int j = 1; j <= c; ++j) {
                scanf("%d %d", &v, &val);
                addedge(i, v, val);
            }
        }
        // 通过枚举第一个点在M区间的最左边和最右边 
        for (int i = 0; i <= M; ++i) {
            lwall = lvl[1]-i, rwall = lwall+M;
            if (lwall < 0) break; // 左边界不可能为负数 
            // 确定好了左右界在进行构边
            build();
            floyd(ret);
        }
        printf("%d\n", ret);
    }
    return 0;    
}