二分+树型 dp hdu5682 zxa and leaf
来源:互联网 发布:林俊杰 睡粉 知乎 编辑:程序博客网 时间:2024/06/07 05:15
传送门:点击打开链接
题意:一棵树n个点,其中有一些点已经有权值,现在给剩下的点安排权值,使得树中相邻两点的之差绝对值的最大值最小。
思路:如果我们首先就想到了二分,那后面很好想了。。
直接二分答案,之后check中,我们随便取1个点为根节点,然后从下向上按拓扑序做树型dp。设SL[u]和SR[u]表示节点u能填的数字的范围
我们从下往上,然后只要判断是否有交集,即有解,我们就能知道当前答案是否可以使用了。
Trick:下次再用G++交用了#prama扩栈的代码剁手。。
#include <map>#include <set>#include <cmath>#include <ctime>#include <stack>#include <queue>#include <cstdio>#include <cctype>#include <bitset>#include <string>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <functional>#define fuck(x) cout<<"["<<x<<"]";#define FIN freopen("input.txt","r",stdin);#define FOUT freopen("output.txt","w+",stdout);#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long LL;typedef pair<int, int> PII;const int MX = 1e5 + 5;const LL INF = 0x3f3f3f3f3f3f3f3fLL;struct Edge { int nxt, v;} E[MX];int Head[MX], erear;void edge_init() { erear = 0; memset(Head, -1, sizeof(Head));}void edge_add(int u, int v) { E[erear].v = v; E[erear].nxt = Head[u]; Head[u] = erear++;}int is[MX], val[MX];LL SL[MX], SR[MX];bool DFS(int u, int f, int x) { if(is[u]) SL[u] = SR[u] = val[u]; else SL[u] = -INF, SR[u] = INF; for(int i = Head[u]; ~i; i = E[i].nxt) { int v = E[i].v; if(v == f) continue; if(!DFS(v, u, x)) return false; if(SL[v] != INF) SL[u] = max(SL[u], SL[v] - x); if(SR[v] != INF) SR[u] = min(SR[u], SR[v] + x); } if(SL[u] > SR[u]) return false; return true;}int solve() { int l = 0, r = 1e9, m; while(l <= r) { m = (l + r) >> 1; if(DFS(1, -1, m)) r = m - 1; else l = m + 1; } return r + 1;}int main() { int T, n, k; //FIN; scanf("%d", &T); while(T--) { edge_init(); scanf("%d%d", &n, &k); for(int i = 1; i <= n; i++) is[i] = 0; for(int i = 1; i <= n - 1; i++) { int u, v; scanf("%d%d", &u, &v); edge_add(u, v); edge_add(v, u); } for(int i = 1; i <= k; i++) { int u, w; scanf("%d%d", &u, &w); is[u] = 1; val[u] = w; } printf("%d\n", solve()); } return 0;}
0 0
- 二分+树型 dp hdu5682 zxa and leaf
- hdu5682 zxa and leaf【dfs+树形dp】
- hdu 5682 zxa and leaf(树形DP+二分)
- HDU 5682 zxa and leaf 二分 树形dp
- hdu 5682 zxa and leaf 二分答案
- HDU 5682:zxa and leaf 二分
- BestCoder #83 1003 zxa and leaf(二分查找/BFS)
- hdu 5682 zxa and leaf (二分+搜索)
- HDU 5682 zxa and leaf
- HDU 5682 zxa and leaf
- BestCoder zxa and set
- 1004 zxa and xor
- BC zxa and set
- BestCoder Round #83 zxa and wifi(一个奇怪的DP题)
- HDU 5680 zxa and set
- HDU 5681 zxa and wifi
- HDU 5683 zxa and xor
- HDU 5680 - zxa and set
- Oracle 查询锁之间的依赖关系
- JS,Jquery获取各种屏幕的宽度和高度
- ubuntu 下配置Source Code Pro字体
- 11. Container With Most Water
- 理解dropout
- 二分+树型 dp hdu5682 zxa and leaf
- angularjs-表单验证
- JS大总结(带实例)
- 消息中间件及WebSphere MQ入门
- hdoj-1789-Doing Homework again
- 键盘按钮keyCode大全
- SpringMVC annotation注解版helloworld
- 北京市工作居住证官方网站js报错不能使用解决办法
- HDU 5680 zxa and set