HDU 5813 Elegant Construction

构造。从a[i]最小的开始放置，例如放置了a[p]，那么还未放置的，还需要建边的那个点 需求量-1，然后把边连起来。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-8;
void File()
{
    freopen("D:\in.txt","r",stdin);
    freopen("D:\out.txt","w",stdout);
}
inline int read()
{
    char c = getchar(); while(!isdigit(c)) c = getchar(); int x = 0;
    while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar();  }
    return x;
}

const int maxn=1000+10;
int T,n,sz;
struct X{int f,num; }s[maxn],ans[maxn*maxn];

bool cmp(X a, X b) {return a.num<b.num; }

int main()
{
    scanf("%d",&T); int cas=1;
    while(T--)
    {
        scanf("%d",&n); sz=0;
        for(int i=1;i<=n;i++) { scanf("%d",&s[i].num); s[i].f=i; }
        sort(s+1,s+1+n,cmp);
        bool fail=0;
        for(int i=1;i<=n;i++)
        {
            if(s[i].num!=0) { fail=1; break; }
            for(int j=i+1;j<=n;j++)
            {
                if(s[j].num==0) continue;
                ans[sz].f=s[i].f; ans[sz].num=s[j].f;  sz++; s[j].num--;
            }
        }

        printf("Case #%d: ",cas++);
        if(fail==1) { printf("No
");  continue; }
        else printf("Yes
");
        printf("%d
",sz);
        for(int i=0;i<sz;i++) printf("%d %d
",ans[i].num,ans[i].f);
    }
    return 0;
}