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 + 10;
const int mod = 1e9 + 7;

int n, m;
int a[11], b[11];

int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d", &n, &m);
        for (int i = 0; i < 3; i++) scanf("%d%d", a + i, b + i);
        LL ans = 0;
        int x, y;
        for (int i = 1; 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 = 0; j < 3; j++) {
                tmp = min(tmp, abs(x - a[j]) + abs(b[j] - y) + 1);
                tmp = min(tmp, abs(x - b[j]) + abs(a[j] - y) + 1);
            }

            //走两条
            for (int j = 0; j < 3; j++) {
                for (int k = 0; k < 3; 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) + 2);
                    //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) + 2);
                    //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) + 2);
                    //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) + 2);
                }
            }

            //走三条
            for (int j = 0; j < 3; j++) {
                for (int k = 0; k < 3; k++) {
                    if (j == k) continue;
                    for (int w = 0; w < 3; 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) + 3);
                        tmp = min(tmp, abs(x - a[j]) + abs(b[j] - a[k]) + abs(b[k] - b[w]) + abs(a[w] - y) + 3);
                        tmp = min(tmp, abs(x - a[j]) + abs(b[j] - b[k]) + abs(a[k] - a[w]) + abs(b[w] - y) + 3);
                        tmp = min(tmp, abs(x - a[j]) + abs(b[j] - b[k]) + abs(a[k] - b[w]) + abs(a[w] - y) + 3);

                        tmp = min(tmp, abs(x - b[j]) + abs(a[j] - a[k]) + abs(b[k] - a[w]) + abs(b[w] - y) + 3);
                        tmp = min(tmp, abs(x - b[j]) + abs(a[j] - a[k]) + abs(b[k] - b[w]) + abs(a[w] - y) + 3);
                        tmp = min(tmp, abs(x - b[j]) + abs(a[j] - b[k]) + abs(a[k] - a[w]) + abs(b[w] - y) + 3);
                        tmp = min(tmp, abs(x - b[j]) + abs(a[j] - b[k]) + abs(a[k] - b[w]) + abs(a[w] - y) + 3);
                    }
                }
            }
            ans += ((LL)i*tmp)%mod;
            ans %= mod;
        }
        printf("%lld
", ans);
    }
    return 0;
}

 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 + 10;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;

int n, m;
int a[11],mat[11][11];

void init() {
    memset(mat, 0x3f, sizeof(mat));
    for (int i = 0; i < 11; i++) mat[i][i] = 0;
}

int main() {
    int tc;
    scanf("%d", &tc);
    while (tc--) {
        init();
        scanf("%d%d", &n, &m);
        for (int i = 0; i < 3; i++) {
            scanf("%d%d", a + i, a + i + 3);
        }
        //新图
        for (int i = 0; i < 6; i++) {
            for (int j = 0; j < 6; j++) {
                mat[i][j] = abs(a[i] - a[j]);
            }
        }
        for (int i = 0; i < 3; i++) {
            mat[i][i + 3] = mat[i + 3][i] = 1;
        }
        //floyd
        for (int k = 0; k < 6; k++) {
            for (int i = 0; i < 6; i++) {
                for (int j = 0; j < 6; j++) {
                    mat[i][j] = min(mat[i][j], mat[i][k] + mat[k][j]);
                }
            }
        }
        LL ans = 0;
        for (int i = 1; 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 = 0; i < 6; i++) {
                for (int j = 0; j < 6; 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
", ans);
    }
    return 0;
}