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  • CodeForces 28D Don't fear, DravDe is kind dp

    主题链接:点击打开链接

    为了让球队后,删除是合法的。也就是说,对于每一个车辆, l+r+c 一样,按l+r+c分类。

    然后dp一下。

    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    #include <math.h>
    #include <set>
    #include <vector>
    #include <map>
    using namespace std;
    #define ll int
    #define N 100005
    struct node{
    	int v,c,l,r;
    	int num;
    }car[N];
    ll n;
    vector<int>G[N], tmp;
    set<int> myset;
    set<int> ::iterator p;
    map<int,int> mymap;
    map<int,int> per;
    map<int,int> sig;
    int pos[N], endpos;
    ll work(ll x){
    	if(G[x].size()==0)return 0;
    	mymap.clear();
    	per.clear();
    	sig.clear();
    	mymap[0] = 0;
    	per[0] = -1;
    	sig[0] = -1;
    	int ans = 0;
    	endpos = -1;
    /*	cout<<"---x:"<<x<<endl;
    	for(int i = 0; i < G[x].size(); i++)cout<<G[x][i]<<" ";
    	puts("---");/**/
    	for(int i = 0; i < G[x].size(); i++) {
    		node now = car[G[x][i]];		
    		if(mymap.find(now.l)==mymap.end())
    			continue;
    	//	cout<<"now: "<<G[x][i]<<endl;
    		if(mymap.find(now.l+now.c)==mymap.end() || mymap[now.l+now.c]<mymap[now.l]+now.v) {
    			mymap[now.l+now.c] = mymap[now.l]+now.v;
    			sig[now.l+now.c] = G[x][i];
    			per[G[x][i]] = sig[now.l];
    		}
    		if(now.r==0 && ans<mymap[now.l+now.c])ans = mymap[now.l+now.c], endpos = G[x][i];
    	}
    	return ans;
    }
    int main(){
    	ll i,j,u,v;
    	while(cin>>n){
    		for(i = 1; i <= n; i++) G[i].clear();
    		myset.clear();
    		mymap.clear();
    		for(i = 1; i <= n; i++)
    			scanf("%d%d%d%d",&car[i].v,&car[i].c,&car[i].l,&car[i].r), car[i].num = i;
    		for(i = 1; i <= n; i++) {
    			myset.insert(car[i].c+car[i].l+car[i].r);
    		}
    		i = 1;
    		for(p = myset.begin(); p!=myset.end(); p++, i++)
    			mymap[*p] = i, pos[i] = *p;
    		for(i = 1; i <= n; i++)
    			G[mymap[car[i].c+car[i].l+car[i].r]].push_back(i);
    		int ans = 0;
    		for(i = 1; i <= n; i++) {
    			int now = work(i);
    			if(now>ans) {
    				ans = now;
    				tmp.clear();
    				u = endpos;
    				while(u!=-1) {
    					tmp.push_back(u);
    					u = per[u];
    				}			
    			}
    		}
    		cout<<tmp.size()<<endl;
    		for(i = tmp.size()-1; i>=0; i--)
    			printf("%d%c",tmp[i],i?' ':'
    ');
    	}
    	return 0;
    }

     

    版权声明:本文博客原创文章,博客,未经同意,不得转载。

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  • 原文地址:https://www.cnblogs.com/mfrbuaa/p/4621662.html
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