POJ 3171 区间最小花费覆盖 （DP+线段树

  1 #include"bits/stdc++.h"
  2 
  3 #define ll long long
  4 #define vl vector<ll>
  5 #define ci(x) scanf("%d",&x)
  6 #define pi(x) printf("%d\n",x)
  7 #define pl(x) printf("%lld\n",x)
  8 #define rep(i, n) for(int i=0;i<n;i++)
  9 using namespace std;
 10 const int NN = 1e6 + 5;
 11 int n,s,t;
 12 struct P{int x,y,s;};
 13 P a[NN];
 14 bool cmp(P a,P b){
 15     return a.y<b.y;
 16 }
 17 const ll INF = 0x3fffffffffffffff;
 18 struct SegMin {
 19     int N;
 20     vl is;vl mul;vl add;
 21     ll init;
 22     ll merge(ll a, ll b) {
 23         return min(a, b);
 24     }
 25     void push(int o, int L, int R, ll m, ll a) {
 26         is[o] = is[o] * m + a;
 27         mul[o] = mul[o] * m;
 28         add[o] = add[o] * m + a;
 29     }
 30 
 31     SegMin(int n, ll init=INF) {
 32         N = 1;
 33         while (N < n) N *= 2;
 34         this->init = init;
 35         is = vl(N * 4, init);
 36         mul = vl(N * 4, 1);
 37         add = vl(N * 4);
 38     }
 39 
 40     SegMin(vl a, ll init=INF) {
 41         int n = a.size();
 42         N = 1;
 43         while (N < n) N *= 2;
 44         this->init = init;
 45         is = vl(N * 2);
 46         mul = vl(N * 2, 1);
 47         add = vl(N * 2);
 48         copy(a.begin(), a.end(), is.begin() + N);
 49         for (int i = N - 1; i > 0; i--) {
 50             is[i] = merge(is[i << 1], is[i << 1 | 1]);
 51         }
 52     }
 53 
 54     void update(int l, int r, ll m, ll a) {
 55         if (l < r) update(1, 0, N, l, r, m, a);
 56     }
 57 
 58     void update(int o, int L, int R, int l, int r, ll m, ll a) {
 59         if (l <= L && R <= r) {
 60             push(o, L, R, m, a);
 61         } else {
 62             int M = (L + R) >> 1;
 63             push(o, L, M, R);
 64             if (l < M) update(o << 1, L, M, l, r, m, a);
 65             if (r > M) update(o << 1 | 1, M, R, l, r, m, a);
 66             is[o] = merge(is[o << 1], is[o << 1 | 1]);
 67         }
 68     }
 69 
 70     void push(int o, int L, int M, int R) {
 71         if (mul[o] != 1 || add[o] != 0) {
 72             push(o << 1, L, M, mul[o], add[o]);
 73             push(o << 1 | 1, M, R, mul[o], add[o]);
 74             mul[o] = 1;
 75             add[o] = 0;
 76         }
 77     }
 78 
 79     ll query(int l, int r) {
 80         if (l < r) return query(1, 0, N, l, r);
 81         return init;
 82     }
 83 
 84     ll query(int o, int L, int R, int l, int r) {
 85         if (l <= L && R <= r) {
 86             return is[o];
 87         } else {
 88             int M = (L + R) >> 1;
 89             push(o, L, M, R);
 90             ll res = init;
 91             if (l < M) res = merge(res, query(o << 1, L, M, l, r));
 92             if (r > M) res = merge(res, query(o << 1 | 1, M, R, l, r));
 93             is[o] = merge(is[o << 1], is[o << 1 | 1]);
 94             return res;
 95         }
 96     }
 97 };
 98 
 99 int main(){
100     ci(n),ci(s),ci(t);//s从1开始
101     s++,t++;
102     int ma=0;
103     for(int i=0;i<n;i++) ci(a[i].x),ci(a[i].y),ci(a[i].s);
104     for(int i=0;i<n;i++) a[i].x++,a[i].y++,ma=max(ma,a[i].y);
105     sort(a,a+n,cmp);
106     SegMin seg(ma+1);
107     seg.update(0,ma+1,0,INF);
108     seg.update(0,s,0,0);
109 
110     for(int i=0;i<n;i++){
111         if(a[i].y<s) continue;
112         int L=a[i].x-1,R=a[i].y;
113         ll res=seg.query(L,R)+a[i].s;
114         res=min(seg.query(R,R+1),res);//与前面的最小值取min
115         seg.update(R,R+1,0,res);
116     }
117     ll ans=seg.query(t,ma+1);
118     if(ans>=INF) puts("-1");//未覆盖到
119     else pl(ans);
120     return 0;
121 }