HDU 5636 Shortest Path

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5636

题解：

1、暴力枚举：

#include<cmath>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;typedef long long LL;
const int maxn = 1e5 + ;
const int mod = 1e9 + ;int n, m;
int a[], b[];int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d", &n, &m);
        for (int i = ; i < ; i++) scanf("%d%d", a + i, b + i);
        LL ans = ;
        int x, y;
        for (int i = ; i <= m; i++) {
            scanf("%d%d", &x, &y);
            //一条新路线都不走
            int tmp = abs(x - y);            //只走一条
            for (int j = ; j < ; j++) {
                tmp = min(tmp, abs(x - a[j]) + abs(b[j] - y) + );
                tmp = min(tmp, abs(x - b[j]) + abs(a[j] - y) + );
            }            //走两条
            for (int j = ; j < ; j++) {
                for (int k = ; k < ; k++) {
                    if (j == k) continue;
                    //x -> j开头 -> j结尾 -> k开头 -> k结尾 -> y
                    tmp = min(tmp, abs(x - a[j]) + abs(b[j] - a[k]) + abs(b[k] - y) + );
                    //x -> j开头 -> j结尾 -> k结尾 -> k开头 -> y
                    tmp = min(tmp, abs(x - a[j]) + abs(b[j] - b[k]) + abs(a[k] - y) + );
                    //x -> j结尾 -> j开头 -> k开头 -> k结尾 -> y
                    tmp = min(tmp, abs(x - b[j]) + abs(a[j] - a[k]) + abs(b[k] - y) + );
                    //x -> j结尾 -> j开头 -> k结尾 -> k开头 -> y
                    tmp = min(tmp, abs(x - b[j]) + abs(a[j] - b[k]) + abs(a[k] - y) + );
                }
            }            //走三条
            for (int j = ; j < ; j++) {
                for (int k = ; k < ; k++) {
                    if (j == k) continue;
                    for (int w = ; w < ; w++) {
                        if (w == j || w == k) continue;
                        tmp = min(tmp, abs(x - a[j]) + abs(b[j] - a[k]) + abs(b[k] - a[w]) + abs(b[w] - y) + );
                        tmp = min(tmp, abs(x - a[j]) + abs(b[j] - a[k]) + abs(b[k] - b[w]) + abs(a[w] - y) + );
                        tmp = min(tmp, abs(x - a[j]) + abs(b[j] - b[k]) + abs(a[k] - a[w]) + abs(b[w] - y) + );
                        tmp = min(tmp, abs(x - a[j]) + abs(b[j] - b[k]) + abs(a[k] - b[w]) + abs(a[w] - y) + );                        tmp = min(tmp, abs(x - b[j]) + abs(a[j] - a[k]) + abs(b[k] - a[w]) + abs(b[w] - y) + );
                        tmp = min(tmp, abs(x - b[j]) + abs(a[j] - a[k]) + abs(b[k] - b[w]) + abs(a[w] - y) + );
                        tmp = min(tmp, abs(x - b[j]) + abs(a[j] - b[k]) + abs(a[k] - a[w]) + abs(b[w] - y) + );
                        tmp = min(tmp, abs(x - b[j]) + abs(a[j] - b[k]) + abs(a[k] - b[w]) + abs(a[w] - y) + );
                    }
                }
            }
            ans += ((LL)i*tmp)%mod;
            ans %= mod;
        }
        printf("%lld\n", ans);
    }
    return ;
}

2、floyd:

对三条新边对应的关键顶点重新建图，跑一遍floyd最短路，对于查询（xi,yi），我们枚举任意两个关键节点（aj,bj），求min(|xi-aj|+len+|bj-yi|)，其中len表示新图里面aj到bj的最短距离。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 typedef long long LL; const int maxn = 1e5 + ;
 const int mod = 1e9 + ;
 const int INF = 0x3f3f3f3f; int n, m;
 int a[],mat[][]; void init() {
     memset(mat, 0x3f, sizeof(mat));
     for (int i = ; i < ; i++) mat[i][i] = ;
 } int main() {
     int tc;
     scanf("%d", &tc);
     while (tc--) {
         init();
         scanf("%d%d", &n, &m);
         for (int i = ; i < ; i++) {
             scanf("%d%d", a + i, a + i + );
         }
         //新图
         for (int i = ; i < ; i++) {
             for (int j = ; j < ; j++) {
                 mat[i][j] = abs(a[i] - a[j]);
             }
         }
         for (int i = ; i < ; i++) {
             mat[i][i + ] = mat[i + ][i] = ;
         }
         //floyd
         for (int k = ; k < ; k++) {
             for (int i = ; i < ; i++) {
                 for (int j = ; j < ; j++) {
                     mat[i][j] = min(mat[i][j], mat[i][k] + mat[k][j]);
                 }
             }
         }
         LL ans = ;
         for (int i = ; i <= m; i++) {
             int x, y;
             scanf("%d%d", &x, &y);
             LL z = abs(x-y);
             //枚举所有情况
             for (int i = ; i < ; i++) {
                 for (int j = ; j < ; j++) {
                     int tmp = abs(x - a[i]) + mat[i][j] + abs(a[j] - y);
                     z = min(z, (LL)tmp);
                 }
             }
             ans += z*i;
             ans %= mod;
         }
         printf("%lld\n", ans);
     }
     return ;
 }