POJ 1659 Frogs' Neighborhood (贪心)

题意：中文题。

析：贪心策略，先让邻居多的选，选的时候也尽量选邻居多的。

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 lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
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;
}
struct Node{
    int id, num;
    bool operator < (const Node &p) const{
        return num > p.num;
    }
};
Node a[15];
int ans[15][15];

int main(){
    int T;  cin >> T;
    while(T--){
        scanf("%d", &n);
        for(int i = 0; i < n; ++i){
            scanf("%d", &a[i].num);
            a[i].id = i;
        }

        bool ok = true;
        memset(ans, 0, sizeof ans);
        for(int i = 0; i < n; ++i){
            sort(a+i, a+n);
            for(int j = i+1; j < n; ++j){
                if(a[i].num && a[j].num){
                    ans[a[i].id][a[j].id] = ans[a[j].id][a[i].id] = 1;
                    --a[i].num;
                    --a[j].num;
                }
                else break;
            }
            if(a[i].num){ ok = false;  break; }
        }

        if(!ok)  puts("NO");
        else {
            puts("YES");
            for(int i = 0; i < n; ++i){
                for(int j = 0; j < n; ++j)
                    if(j)  printf(" %d", ans[i][j]);
                    else printf("%d", ans[i][j]);
                printf("\n");
            }
        }
        if(T)  puts("");
    }
    return 0;
}