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贪心要注意cow按照maxSPFA从小到大排序,bottle按照SPFA从小到大排序/* * Created: 2016年03月27日 13时08分15秒 星期日 * Author: Akrusher * */ #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <ctime> #include <iostream> #include <algorithm> #include <string> #include <vector> #include <deque> #include <list> #include <set> #include <map> #include <stack> #include <queue> #include <numeric> #include <iomanip> #include <bitset> #include <sstream> #include <fstream> using namespace std; #define rep(i,a,n) for (int i=a;i<n;i++) #define per(i,a,n) for (int i=n-1;i>=a;i--) #define in(n) scanf("%d",&(n)) #define in2(x1,x2) scanf("%d%d",&(x1),&(x2)) #define inll(n) scanf("%I64d",&(n)) #define inll2(x1,x2) scanf("%I64d%I64d",&(x1),&(x2)) #define inlld(n) scanf("%lld",&(n)) #define inlld2(x1,x2) scanf("%lld%lld",&(x1),&(x2)) #define inf(n) scanf("%f",&(n)) #define inf2(x1,x2) scanf("%f%f",&(x1),&(x2)) #define inlf(n) scanf("%lf",&(n)) #define inlf2(x1,x2) scanf("%lf%lf",&(x1),&(x2)) #define inc(str) scanf("%c",&(str)) #define ins(str) scanf("%s",(str)) #define out(x) printf("%d\n",(x)) #define out2(x1,x2) printf("%d %d\n",(x1),(x2)) #define outf(x) printf("%f\n",(x)) #define outlf(x) printf("%lf\n",(x)) #define outlf2(x1,x2) printf("%lf %lf\n",(x1),(x2)); #define outll(x) printf("%I64d\n",(x)) #define outlld(x) printf("%lld\n",(x)) #define outc(str) printf("%c\n",(str)) #define pb push_back #define mp make_pair #define fi first #define se second #define SZ(x) ((int)(x).size()) #define mem(X,Y) memset(X,Y,sizeof(X)); typedef vector<int> vec; typedef long long ll; typedef pair<int,int> P; const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1}; const int INF=0x3f3f3f3f; const ll mod=1e9+7; ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} const bool AC=true; struct point{ int a; int b; }; bool cmp(point x,point y){ return x.b<y.b; } bool cmp2(point x,point y){ return x.a<y.a; } point x[2506]; point y[2506]; int main() { int n,l,ans; in2(n,l); rep(i,0,n) in2(x[i].a,x[i].b); rep(i,0,l) in2(y[i].a,y[i].b); sort(x,x+n,cmp); sort(y,y+l,cmp2); ans=0; rep(i,0,n){ rep(j,0,l){ if(y[j].a>=x[i].a&&y[j].a<=x[i].b&&y[j].b>0){ y[j].b--;ans++; break; } } } out(ans); return 0; } Followed by:
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