UPCOJ 4199

链接:

  http://exam.upc.edu.cn/problem.php?id=4199

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题目：

题目描述

NEUACM team holds summer training every year. This year n students participate in the training, and we provide m computers. Everyone has a suitability (0<=suitability<=1000) to every computer.
Now our coach needs to draw up a plan, which satisfies:
1. every student uses one computer, and a computer can be used by no more than one student;
2. maximize the minimum suitability, which means if the answer is w, then everyone’s suitability to his computer shouldn’t be less than w.

Hint: huge input, please use fast IO.

输入

The first line contains n and m (1<=n<=m<=500), indicating the number of students and computers.

Next n lines describes every student’s suitability to every computer, number in ith line and jth row means student[i]’s suitability to computer[j].

输出

A single number w in one line, which is described before.

样例输入

1 2
1 10
2 3
10 1 2
2 1 1

样例输出

10
2

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题意：

  给你一个n×m的图mp，mp[i][j]代表第i个学生对第j台电脑的适配程度，每台电脑只能由一个人使用，问在所有人都有电脑用的情况下，所有人里对当前电脑的适配值的最小的那个人的适配值的最大值。

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思路：

  直接二分答案，也就是每个人可以成功适配的最小适配值，不断让这个值增大，直到无法完成二分图匹配。

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实现：

#include <iostream>#include <algorithm>#include <set>#include <string>#include <vector>#include <queue>#include <map>#include <stack>#include <list>#include <iomanip>#include <functional>#include <sstream>#include <cstdio>#include <cstring>#include <cmath>#include <cctype>//#define read read()#define edl putchar('\n')#define ll long long#define clr(a,b) memset(a,b,sizeof a)#define rep(i,m,n) for(int i=m ; i<=n ; i++)#define fep(i,n) for(int i=0 ; i<n ; i++)//inline int read{ int x=0;char c=getchar();while(c<'0' || c>'9')c=getchar();while(c>='0' && c<='9'){ x=x*10+c-'0';c=getchar(); }return x;}namespace FastIO {    const int SIZE = 1 << 16;    char buf[SIZE], obuf[SIZE], str[60];    int bi = SIZE, bn = SIZE, opt;    int read(char *s) {        while (bn) {            for (; bi < bn && buf[bi] <= ' '; bi++);            if (bi < bn) break;            bn = fread(buf, 1, SIZE, stdin);            bi = 0;        }        int sn = 0;        while (bn) {            for (; bi < bn && buf[bi] > ' '; bi++) s[sn++] = buf[bi];            if (bi < bn) break;            bn = fread(buf, 1, SIZE, stdin);            bi = 0;        }        s[sn] = 0;        return sn;    }    bool read(int& x) {        int n = read(str), bf;        if (!n) return 0;        int i = 0; if (str[i] == '-') bf = -1, i++; else bf = 1;        for (x = 0; i < n; i++) x = x * 10 + str[i] - '0';        if (bf < 0) x = -x;        return 1;    }};#define read(x) FastIO::read(x)using namespace std;const int maxn = 1007;int n, m, mp[maxn][maxn], link[maxn];bool vis[maxn];bool dfs(int u, int minn) {    fep(i,m) {        int v = i+n;        if(vis[v] || mp[u][v] < minn) continue;        vis[v] = true;        if(link[v] == -1 || dfs(link[v], minn)) {            link[u] = v;            link[v] = u;            return true;        }    }    return false;}bool match(int minn) {    clr(link, -1);    int ans = 0;    fep(i,n)        if(link[i] == -1) {            clr(vis, 0);            if(dfs(i, minn)) ans++;        }    return ans == n;}int main() {#ifndef ONLINE_JUDGE    freopen("../in.txt", "r", stdin);//    printf("init! init! init!\n");#endif    while(read(n)&&read(m)) {        fep(i,n) fep(j,m) read(mp[i][n+j]), mp[n+j][i] = mp[i][n+j];        int l=0, r=1000, mid, ans;        while(l<=r) {            mid = l+r>>1;            if(match(mid)) {                ans = mid;                l = mid+1;            } else r = mid-1;        }        printf("%d\n", ans);    }    return 0;}