UVALive

题目链接

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=5449

题意

给出一个n
然后有n行
每行给出两个数
这两个数之间可以用三种操作 分别是 + - *

如果这n对数通过操作后 得到的结果都是不同的，那么这个方案就是符合条件的 输出任意一种可行方案

如果存在相同结果，那么就是不可行的

思路
可以把N对数看成一个点，把一对数通过三种操作后得到的结果也看成一个点，将他们之间连一条边

然后二分匹配 如果最大匹配数 == n 那么就有可行解

输出就好了

linker[] 里存放的就是匹配的方案

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <list>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a, b) memset(a, (b), sizeof(a))
#define pb push_back
#define bug puts("***bug***");
#define fi first
#define se second
//#define bug 
//#define gets gets_s

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <string, int> psi;
typedef pair <string, string> pss;
typedef pair <double, int> pdi;

const double PI = acos(-1.0);
const double EI = exp(1.0);
const double eps = 1e-8;

const int INF = 0x3f3f3f3f;
const int maxn = 7e3 + 5e2 + 10;
const int MOD = 6;                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        


const int MAXN = 7510;//点数的最大值
const int MAXM = 50010;//边数的最大值
struct Edge
{
    int to, next;
}edge[MAXM];
int head[MAXN], tot;
void init()
{
    tot = 0;
    memset(head, -1, sizeof(head));
}
void addedge(int u, int v)
{
    edge[tot].to = v; edge[tot].next = head[u];
    head[u] = tot++;
}
int linker[MAXN];
bool used[MAXN];
int uN;
bool dfs(int u)
{
    for (int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].to;
        if (!used[v])
        {
            used[v] = true;
            if (linker[v] == -1 || dfs(linker[v]))
            {
                linker[v] = u;
                return true;
            }
        }
    }
    return false;
}
int hungary()
{
    int res = 0;
    memset(linker, -1, sizeof(linker));
    for (int u = 0; u < uN; u++)//点的编号0~uN-1
    {
        memset(used, false, sizeof(used));
        if (dfs(u))res++;
    }
    return res;
}

ll A[MAXN], B[MAXN], C[MAXN];

int main()
{
    int n;
    while (scanf("%d", &n) != EOF)
    {
        init();
        map <ll, int> mp;
        int pos = 0;
        for (int i = 0; i < n; i++)
        {
            scanf("%lld%lld", &A[i], &B[i]);
            ll num;
            num = A[i] + B[i];
            if (mp[num] == 0)
            {
                C[pos] = num;
                mp[num] = ++pos;
            }
            addedge(mp[num] - 1, i);
            num = A[i] - B[i];
            if (mp[num] == 0)
            {
                C[pos] = num;
                mp[num] = ++pos;
            }
            addedge(mp[num] - 1, i);
            num = A[i] * B[i];
            if (mp[num] == 0)
            {
                C[pos] = num;
                mp[num] = ++pos;
            }
            addedge(mp[num] - 1, i);
        }
        uN = pos;
        if (hungary() < n)
            puts("impossible");
        else
        {
            for (int i = 0; i < n; i++)
            {
                ll num = C[linker[i]];
                char c;
                if (A[i] + B[i] == num)
                    c = '+';
                else if (A[i] - B[i] == num)
                    c = '-';
                else
                    c = '*';
                printf("%lld %c %lld = %lld
", A[i], c, B[i], num);
            }
        }
    }
}