Gym

题目链接：https://vjudge.net/contest/181019#problem/F

题意：给定一个n*n的矩阵，从左下至右上平行于对角线的斜线共2*n-1条（包括对角线），在每条斜线上选取一个出现的值，要求这2*n-1条斜线选取的值互不相同，求是否能选取，若不能，输出NO，若能，输出YES并按1~2*n-1的顺序输出每条斜线上选取的值。

思路：从左下至右上为这些斜线编号为1~2*n-1，将编号x与斜线x上出现的数值进行连线，求出最大匹配数是否为2*n-1即可。注意数值较大，可以为每个数赋一个较小的id值放置vis数组越界。

代码如下：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<cstdlib>#include<sstream>#include<deque>#include<stack>#include<set>#include<map>using namespace std;typedef long long ll;typedef unsigned long long ull;const double eps = 1e-6;const int  maxn = 1000 + 20;const int  maxt = 300 + 10;const int mod = 10;const int dx[] = {1, -1, 0, 0};const int dy[] = {0, 0, -1, 1};const int Dis[] = {-1, 1, -5, 5};const int inf = 0x3f3f3f3f;const int MOD = 1000;int n, m, k;int d[maxn][maxn];int match[100000 + 10];//因match数组开小WA一次bool vis[100000 + 10];//因vis数组开太小RE一次vector<int> g[maxn];map<int, int> id;map<int, int> pos;int ans[maxn];bool dfs(int u){    int v;    int len = g[u].size();    for(int i = 0; i < len; ++i){        v = g[u][i];        if(vis[v]) continue;        vis[v] = true;        if(match[v] == -1 || dfs(match[v])){            match[v] = u;            ans[u] = v;//记录每条斜线的当前匹配值            return true;        }    }    return false;}int main(){    scanf("%d", &n);    int xx = 610;    id.clear(); pos.clear();    memset(d, 0, sizeof d);    for(int i = 0; i < maxn; ++i) g[i].clear();    for(int i = 1; i <= n; ++i){        for(int j = 1; j <= n; ++j){            scanf("%d",&d[i][j]);            if(id.find(d[i][j]) == id.end()){//d[i][j]的值太大，根据输入顺序为其编号，防止RE                id[d[i][j]] = xx;                pos[xx++] = d[i][j];            }        }    }    int t = n;    int num = 1, x = 2 * n;    while(t >= 1){//遍历每一条斜线，从左下至右上id编号依次为1~2*n-1，将斜线上存在的数值id值与斜线id值连线        for(int i = t, j = 1; i <= n; ++i,++j){            if(t != 1){                g[num].push_back(id[d[i][j]]);                g[x - num].push_back(id[d[j][i]]);            }            else{                g[num].push_back(id[d[i][j]]);            }        }        ++num; --t;    }    int cnt = 0;    bool ok = true;    memset(ans, 0, sizeof ans);    memset(match, -1, sizeof match);    for(int i = 1; i < x; ++i){//为每一条斜线寻找匹配值        memset(vis, false, sizeof vis);        if(dfs(i)){            ++cnt;        }        else{//若有某条斜线无匹配值，直接退出            ok = false; break;        }    }    if(!ok || cnt != 2 * n - 1){        printf("NO\n"); return 0;    }    printf("YES");    for(int i = 1; i < x; ++i){        printf(" %d", pos[ans[i]]);    }    printf("\n");    return 0;}/*51000000000 1000000000 1000000000 900000000 2100000001000000000 200000000 200000000 800000000 7000000001000000000 200000000 300000000 300000000 7000000001000000000 200000000 600000000 400000000 700000000700000000  700000000  700000000 00000000   700000000//1 2 6 4 3 8 941 1 1 91 2 2 81 2 4 41 2 6 551 2 3 4 56 7 8 9 1011 12 13 14 1516 17 18 19 2021 22 23 24 2531 1 12 2 43 1 1*/