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  • [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;
    }
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  • 原文地址:https://www.cnblogs.com/shandongs1/p/8253498.html
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