• 【NOIP2012】开车旅行(倍增+STL)


    【NOIP2012】开车旅行
    纪念一下这个大毒瘤
    先用multiset处理一下在每个城市时小A和小B的下一个到达的。
    一个一个地跳肯定会超时,所以考虑倍增。
    f[i][A/B][k]表示从城市i出发,A/B先开车,走(2^k)天所到达的城市。
    fa[i][A/B][k]表示从城市i出发,A/B先开车,走(2^k)天小A走的路程。
    fb[i][A/B][k]表示从城市i出发,A/B先开车,走(2^k)天小B走的路程。
    初始化:

     //0表示小A, 1表示小B higha表示小A到达的下一个城市,highb同理
    f[i][0][0] = higha; f[i][1][0] = highb;
    fa[i][0][0] = abs(h[i] - h[higha]);
    fb[i][1][0] = abs(h[i] - h[highb]);
    

    转移

    //这里j表示天数的幂,即2^j天,k表示小A或小B
    for(int j = 1;j <= 23; j++) {
          for(int k = 0; k < 2; k++) {
    	      if(j == 1) {//走两天时单独处理
    	              f[i][k][1] = f[f[i][k][0]][1 - k][0];
    		      fa[i][k][1] = fa[i][k][0] + fa[f[i][k][0]][1 - k][0];
    		      fb[i][k][1] = fb[i][k][0] + fb[f[i][k][0]][1 - k][0];
    	      } else {
    	            f[i][k][j] = f[f[i][k][j - 1]][k][j - 1];
    	            fa[i][k][j] = fa[i][k][j - 1] + fa[f[i][k][j - 1]][k][j - 1];
    	            fb[i][k][j] = fb[i][k][j - 1] + fb[f[i][k][j - 1]][k][j - 1];
    	      }
          }
    }
    

    完整代码:

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <set>
    using namespace std;
    typedef long long lld;
    const int N = 100100;
    const int inf = 2e9;
    int a[N], n, h[N];
    int f[N][2][25], fa[N][2][25], fb[N][2][25];//0 : A, 1 : B
    //A 次近 B 最近
    struct node {
    	int id, high;
    	friend bool operator < (node a, node b) {//重载运算符由小到大排序
    		return a.high < b.high;
    	}
    };
    multiset<node> s;
    void get_next() {
    	h[0] = inf; h[n + 1] = -inf;
    	node st;
    	st.id = 0, st.high = inf;
    	s.insert(st); s.insert(st);
    	st.id = n + 1, st.high = -inf;
    	s.insert(st); s.insert(st);
    	set<node>::iterator q;
    	for(int i = n; i >= 1; i--) {
    		int higha, highb;
    		node city; city.high = h[i], city.id = i;
    		s.insert(city);
    		q = s.lower_bound(city);
    		q--;
    		int per = (*q).id, perh = (*q).high;
    		q++, q++;
    		int nxt = (*q).id, nxth = (*q).high;
    		q--;
    		if(abs(perh - h[i]) <= abs(nxth - h[i])) { //per min
    			highb = per;
    			q--, q--;
    			if(abs(nxth - h[i]) < abs(h[i] - (*q).high)) {
    				higha = nxt;
    			} else higha = (*q).id;
    		} else { //nxt min
    			highb = nxt;
    			q++, q++;
    			if(abs(perh - h[i]) <= abs(h[i] - (*q).high)) {
    				higha = per;
    			} else higha = (*q).id;
    		}
    		f[i][0][0] = higha; f[i][1][0] = highb;
    		fa[i][0][0] = abs(h[i] - h[higha]);
    		fb[i][1][0] = abs(h[i] - h[highb]);
    		for(int j = 1;j <= 23; j++) {
    			for(int k = 0; k < 2; k++) {
    				if(j == 1) {
    					f[i][k][1] = f[f[i][k][0]][1 - k][0];
    					fa[i][k][1] = fa[i][k][0] + fa[f[i][k][0]][1 - k][0];
    					fb[i][k][1] = fb[i][k][0] + fb[f[i][k][0]][1 - k][0];
    				} else {
    					f[i][k][j] = f[f[i][k][j - 1]][k][j - 1];
    					fa[i][k][j] = fa[i][k][j - 1] + fa[f[i][k][j - 1]][k][j - 1];
    					fb[i][k][j] = fb[i][k][j - 1] + fb[f[i][k][j - 1]][k][j - 1];
    				}
    			}
    		}
    	}
    }
    int xa, xb;
    void calc(int start, int x) {
    	int st = start;
    	xa = 0, xb = 0;
    	for(int i = 23; i >= 0; i--) {
    		if(f[st][0][i] && xa + xb + fa[st][0][i] + fb[st][0][i] <= x) {
    			xa += fa[st][0][i]; xb += fb[st][0][i];
    			st = f[st][0][i];
    		}
    	}
    	return ;
    }
    void solve1() {
    	int x0; scanf("%d", &x0);
    	double ratio = inf * 1.0; int ans = 0;
    	for(int i = 1; i <= n; i++) {
    		calc(i, x0);
    		if(xb == 0) {
    			xb = 1; xa = inf;
    		}
    		if((1.0 * double(xa) / double(xb)) < ratio) {
    			ratio = double(xa) / double(xb); ans = i;
    		} else if(double(xa) / double(xb) == ratio) {
    			if(h[i] > h[ans]) ans = i;
    		}
    	}
    	printf("%d
    ", ans);
    }
    void solve2() {
    	int m; scanf("%d", &m);
    	for(int i = 1, x0, st; i <= m; i++) {
    		scanf("%d%d", &st, &x0);
    		calc(st, x0);
    		printf("%d %d
    ", xa, xb);
    	}
    }
    int main() {
    //	freopen("data.in", "r", stdin);
    //	freopen("data.out", "w", stdout);
    	scanf("%d", &n);
    	for(int i = 1; i <= n; i++) {
    		scanf("%d", &h[i]);
    	}
    	get_next();
    	solve1();
    	solve2();
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/mcggvc/p/13796053.html
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