Poj--3171（线段树，DP）

2014-12-07 12:31:31

思路：经典的区间覆盖DP，dp[i]表示[M,i]被覆盖的最小费用，转移方程：dp[i] = min(dp[i],dp[k] + fee[i]) (T1[i] - 1 <= k <= T2[i] - 1)

　　dp值用线段树维护，优化到nlogn。注意：题目中的时间有可能为0，所以如果以1为根建树的话，要把所有时间+1再处理。

  1 /*************************************************************************
  2     > File Name: 3171.cpp
  3     > Author: Nature
  4     > Mail: 564374850@qq.com 
  5     > Created Time: Sun 07 Dec 2014 01:42:24 AM CST
  6 ************************************************************************/
  7 
  8 #include <cstdio>
  9 #include <cstring>
 10 #include <cstdlib>
 11 #include <cmath>
 12 #include <vector>
 13 #include <map>
 14 #include <set>
 15 #include <stack>
 16 #include <queue>
 17 #include <iostream>
 18 #include <algorithm>
 19 using namespace std;
 20 #define lp (p << 1)
 21 #define rp (p << 1|1)
 22 #define getmid(l,r) (l + (r - l) / 2)
 23 #define MP(a,b) make_pair(a,b)
 24 typedef long long ll;
 25 const ll INF = 1e16;
 26 const int maxn = 90000;
 27 
 28 int N,M,E;
 29 ll dp[maxn];
 30 
 31 struct node{
 32     int st,ed,sl;
 33 }t[100010];
 34 
 35 struct SegTree{
 36     ll t[maxn << 2];
 37     void Build(int p,int l,int r){
 38         t[p] = INF;
 39         if(l == r)
 40             return;
 41         int mid = getmid(l,r);
 42         Build(lp,l,mid);
 43         Build(rp,mid + 1,r);
 44     }
 45     void Update(int a,ll c,int p,int l,int r){
 46         if(l == r){
 47             t[p] = min(t[p],c);
 48             return;
 49         }
 50         int mid = getmid(l,r);
 51         if(a <= mid) Update(a,c,lp,l,mid);
 52         else Update(a,c,rp,mid + 1,r);
 53         t[p] = min(t[lp],t[rp]);
 54     }
 55     ll Query(int a,int b,int p,int l,int r){
 56         if(a <= l && r <= b){
 57             return t[p];
 58         }
 59         int mid = getmid(l,r);
 60         ll res = INF;
 61         if(a <= mid) res = min(res,Query(a,b,lp,l,mid));
 62         if(b > mid) res = min(res,Query(a,b,rp,mid + 1,r));
 63         return res;
 64     }
 65 }ST;
 66 
 67 bool cmp(node a,node b){
 68     return a.ed < b.ed;
 69 }
 70 
 71 int main(){
 72     scanf("%d%d%d",&N,&M,&E);
 73     for(int i = 0; i < maxn; ++i)
 74         dp[i] = INF;
 75     ST.Build(1,1,maxn);
 76     ++M;
 77     ++E;
 78     for(int i = 1; i <= N; ++i){
 79         scanf("%d%d%d",&t[i].st,&t[i].ed,&t[i].sl);
 80         ++t[i].st;
 81         ++t[i].ed;
 82     }
 83     sort(t + 1,t + N + 1,cmp);
 84     for(int i = 1; i <= N; ++i){
 85         if(t[i].st > E || t[i].ed < M)
 86             continue;
 87         if(t[i].st <= M){
 88             dp[t[i].ed] = min(dp[t[i].ed],(ll)t[i].sl);
 89         }
 90         else{
 91             int bot = max(t[i].st - 1,M),top = min(E,max(t[i].ed - 1,M));
 92             ll tmin = ST.Query(bot,top,1,1,maxn);
 93             dp[t[i].ed] = min(dp[t[i].ed],tmin + t[i].sl);
 94         }
 95         ST.Update(t[i].ed,dp[t[i].ed],1,1,maxn);
 96     }
 97     if(dp[E] >= INF) printf("-1\n");
 98     else printf("%I64d\n",dp[E]);
 99     return 0;
100 }

 

posted @ 2014-12-07 12:36  Naturain  阅读(162)  评论(0)  编辑  收藏  举报