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题解:
将一边排序过后就是裸的逆序题了
这里有2种方法
1.add()往下更新,sum()往上更新 求逆序sum(a+1);
2.add)往上更新,sum()往下更新 求逆序sum(maxn)-sum(a);
代码:
1.
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<map> #include<set> using namespace std; #define pb push_back #define mp make_pair #define se second #define fs first #define LL long long #define CLR(x) memset(x,0,sizeof x) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 typedef pair<int,int> P; const double eps=1e-9; const int maxn=20100; const int N=110; const int mod=1e9+7; const int INF=1e9; int T,n,m,k; LL c[maxn]; P p[maxn*100]; int lowbit(int x){return x&-x;} void add(int x){ while(x){ c[x]++; x-=lowbit(x); } } LL sum(int x){ LL cnt=0; while(x<maxn){ cnt+=c[x]; x+=lowbit(x); } return cnt; } int main(){ int Case=1; scanf("%d",&T); while(T--){ CLR(c); scanf("%d%d%d",&n,&m,&k); for(int i=1;i<=k;i++) scanf("%d%d",&p[i].fs,&p[i].se); sort(p+1,p+k+1); LL Sum=0; for(int i=1;i<=k;i++){ Sum+=sum(p[i].se+1); //cout<<Sum<<endl; add(p[i].se); } printf("Test case %d: %lld\n",Case++,Sum); } return 0; }
2.
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<map> #include<set> using namespace std; #define pb push_back #define mp make_pair #define se second #define fs first #define LL long long #define CLR(x) memset(x,0,sizeof x) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 typedef pair<int,int> P; const double eps=1e-9; const int maxn=20100; const int N=110; const int mod=1e9+7; const int INF=1e9; int T,n,m,k; LL c[maxn]; P p[maxn*100]; int lowbit(int x){return x&-x;} void add(int x){ while(x<maxn){ c[x]++; x+=lowbit(x); } } LL sum(int x){ LL cnt=0; while(x){ cnt+=c[x]; x-=lowbit(x); } return cnt; } int main(){ int Case=1; scanf("%d",&T); while(T--){ CLR(c); scanf("%d%d%d",&n,&m,&k); for(int i=1;i<=k;i++) scanf("%d%d",&p[i].fs,&p[i].se); sort(p+1,p+k+1); LL Sum=0; for(int i=1;i<=k;i++){ Sum+=sum(maxn)-sum(p[i].se); //cout<<Sum<<endl; add(p[i].se); } printf("Test case %d: %lld\n",Case++,Sum); } return 0; }
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原文地址：http://www.cnblogs.com/byene/p/5897889.html
