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HDU - 2795 Billboard (线段树)
题意:有n个1*wi的传单要放在一个h*w的宣传板上,在尽量往上放的基础上尽量往左放,问每个传单放在第几行,若无法放上,则输出-1。
分析:二分,用线段树维护,若前半区间所有行的最大值大于等于wi,那么继续在前半区间搜索,否则搜索后半区间。
#include<cstdio>#include<cstring>#include<cstdlib>#include<cctype>#include<cmath>#include<iostream>#include<sstream>#include<iterator>#include<algorithm>#include<string>#include<vector>#include<set>#include<map>#include<stack>#include<deque>#include<queue>#include<list>#define lowbit(x) (x & (-x))const double eps = 1e-8;inline int dcmp(double a, double b){ if(fabs(a - b) < eps) return 0; return a > b ? 1 : -1;}typedef long long LL;typedef unsigned long long ULL;const int INT_INF = 0x3f3f3f3f;const int INT_M_INF = 0x7f7f7f7f;const LL LL_INF = 0x3f3f3f3f3f3f3f3f;const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};const int MOD = 1e9 + 7;const double pi = acos(-1.0);const int MAXN = 200000 + 10;const int MAXT = 10000 + 10;using namespace std;int input[MAXN << 2];int _max[MAXN << 2];int h, w, n;void update(int id, int L, int R, int v){ _max[id] = v; if(L == R) return; int mid = L + (R - L) / 2; update(id << 1, L, mid, v); update(id << 1 | 1, mid + 1, R, v);}int solve(int id, int L, int R, int v){ if(L == R){ _max[id] -= v; return R; } int mid = L + (R - L) / 2; int ans = 0; if(_max[id << 1] >= v){ ans = solve(id << 1, L, mid, v); } else{ ans = solve(id << 1 | 1, mid + 1, R, v); } _max[id] = max(_max[id << 1], _max[id << 1 | 1]); return ans;}int main(){ while(scanf("%d%d%d", &h, &w, &n) == 3){ memset(input, 0, sizeof input); memset(_max, 0, sizeof _max); for(int i = 1; i <= n; ++i){ scanf("%d", &input[i]); } h = min(h, n); update(1, 1, h, w); for(int i = 1; i <= n; ++i){ if(input[i] > w){ printf("-1\n"); continue; } if(input[i] > _max[1]){ printf("-1\n"); continue; } printf("%d\n", solve(1, 1, h, input[i])); } } return 0;}
HDU - 2795 Billboard (线段树)
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