URAL 2069 Hard Rock (最短路)

题意：给定 n + m 个街道，问你从左上角走到右下角的所有路的权值最小的中的最大的。

析：我们只要考虑几种情况就好了，先走行再走列和先走列再走行差不多。要么是先横着，再竖着，要么是先横再竖再横，要么是先横再竖再横再竖，全考虑一下就好了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#define frer freopen("in.txt", "r", stdin)

#define frew freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1e5 + 5;

const int mod = 1e9 + 7;

const char *mark = "+-*";

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline int Min(int a, int b){ return a < b ? a : b; }

inline int Max(int a, int b){ return a > b ? a : b; }

inline LL Min(LL a, LL b){ return a < b ? a : b; }

inline LL Max(LL a, LL b){ return a > b ? a : b; }

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

int row[maxn], col[maxn];

int solve(int *a, int r, int *b, int c){

    int ans1 = min(a[0], b[c]);

    int ans2 = min(a[0], a[r]);

    int cnt = 0;

    for(int i = 0; i <= c; ++i)

        cnt = max(cnt, b[i]);

    ans2 = min(ans2, cnt);

    int ans3 = min(a[0], b[c]);

    int rr = 0;

    for(int i = 0; i <= r; ++i)

        rr = max(rr, a[i]);

    ans2 = min(ans2, min(cnt, rr));

    return max(ans1, max(ans2, ans3));

}

int main(){

    while(scanf("%d %d", &m, &n) == 2){

        for(int i = 0; i < m; ++i)

            scanf("%d", &col[i]);

        for(int j = 0; j < n; ++j)

            scanf("%d", &row[j]);

        int ans = solve(row, n-1, col, m-1);

        ans = max(ans, solve(col, m-1, row, n-1));

        printf("%d\n", ans);

    }

    return 0;

}