[Codevs] 一塔湖图

http://codevs.cn/problem/1024/

floyd 走起

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

#define inf 0x7ffffff
#define gc getchar()

int x[10], y[10], n, m, t, k, x1, y1, x2, y2, a, b, f[120][120];

inline int read(){
    int x = 0; char c = gc;
    while(c < '0' || c > '9') c = gc;
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc;
    return x;
}

int main()
{
    n = read(); m = read(); t = read(); k = read();
    for(int i = 1; i <= n; i ++) x[i] = read();
    for(int i = 1; i <= m; i ++) y[i] = read();
    for(int i = 1; i <= n * m; i ++)
        for(int j = 1; j <= n * m; j ++)
            if(i == j) f[i][j] = 0; else f[i][j] = inf;
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= m; j ++){
            if(i > 1) {a = i - 1; b = j; f[(j - 1) * n + i][(b - 1) * n + a] = x[i] - x[i - 1];}
            if(j > 1) {a = i; b = j - 1; f[(j - 1) * n + i][(b - 1) * n + a] = y[j] - y[j - 1];}
            if(i < n) {a = i + 1; b = j; f[(j - 1) * n + i][(b - 1) * n + a] = x[i + 1] - x[i];}
            if(j < m) {a = i; b = j + 1; f[(j - 1) * n + i][(b - 1) * n + a] = y[j + 1] - y[j];}
        }
    for(int i = 1; i <= t; i ++){
        x1 = read(); y1 = read(); x2 = read(); y2 = read();
        f[(y1 - 1) * n + x1][(y2 - 1) * n + x2] = inf;
        f[(y2 - 1) * n + x2][(y1 - 1) * n + x1] = inf;
    }
    for(int i = 1; i <= k; i ++) {
        x1 = read(); x2 = read(); y1 = read(); y2 = read();
        for(a = x1; a <= x2 - 1; a ++)
            for(b = y1 + 1; b <= y2 - 1; b ++) {
                f[(b - 1) * n + a][(b - 1) * n + a + 1] = inf;
                f[(b - 1) * n + a + 1][(b - 1) * n + a] = inf;
            }
        for(b = y1; b <= y2 - 1; b ++)
            for( a = x1 + 1; a <= x2 -1; a ++) {
                f[(b - 1) * n + a][b * n + a] = inf;
                f[b * n + a][(b - 1) * n + a] = inf;
            }
    }
    for(int k = 1; k <= n * m; k ++)
        for(int i = 1; i <= n * m; i ++)
            for(int j = 1; j <= n * m; j ++)
                f[i][j] = min(f[i][k] + f[k][j], f[i][j]);
    x1 = read(); y1 = read(); x2 = read(); y2 = read();
    printf("%d", f[(y1 - 1) * n + x1][(y2 - 1) * n + x2]);
    return 0;
}

 

posted @ 2018-01-09 20:22  xayata  阅读(167)  评论(0编辑  收藏  举报