贪心——HDU 5813 ( 2016 Multi-University Training Contest 7 1005 )
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题目链接:
http://acm.hdu.edu.cn/showproblem.php?pid=5813分析:
N代表N个城市,每个城市要求只与Ai 个城市相连接,且最后的路不能成环。若满足就输出Yes和每条路否则输出NO题解:
先把给出的要求连接城市数排序,然后遍历一遍,若Ai 大于i-1那么不成立,再遍历一遍,每个城市只连接它前面的城市参考代码:
#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <queue>#include <map>#include <set>#include <vector>#include <list>#define LL long long#define eps 1e-8#define maxn 1010#define mod 100000007#define inf 0x3f3f3f3f#define mid(a,b) ((a+b)>>1)using namespace std;int n;typedef pair<int,int> pii;pii num[maxn];int main(){ int t, ca = 1; cin >> t; while(t--) { scanf("%d", &n); for(int i=1; i<=n; i++) { int x; scanf("%d", &x); num[i] = make_pair(x, i); } sort(num+1, num+1+n); int flag = 1; int cnt = 0; for(int i=1; i<=n; i++) { if(num[i].first >= i) { flag = 0; break; } cnt += num[i].first; } if(!flag) { printf("Case #%d: No\n", ca++); continue; } printf("Case #%d: Yes\n", ca++); printf("%d\n", cnt); for(int i=1; i<=n; i++) { for(int j=1; j<=num[i].first; j++) { printf("%d %d\n", num[i].second, num[j].second); } } } return 0;}
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