hdu5550Game Rooms

链接:http://acm.hdu.edu.cn/showproblem.php?pid=5550

题意:有n层楼,每层有p[i]个人喜欢网球t[i]个人喜欢游泳,每层楼只能建一个娱乐场所,没有场所的人要去其他楼,求建好娱乐场所后所有人到离他最近的喜欢的娱乐场所的总和最小。

分析:设dp[i][j][0/1]表示第i层建0/1,j

代码:

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=4010;
const int mod=100000000;
const int MOD1=1000000007;
const int MOD2=1000000009;
const double EPS=0.00000001;
typedef long long ll;
const ll MOD=1000000007;
const int MAX=1000000010;
const ll INF=1ll<<55;
const double pi=acos(-1.0);
typedef double db;
typedef unsigned long long ull;
ll x[N],y[N],dp[2][N][2];
ll sumx[N],sumy[N],ssumx[N],ssumy[N];
void init(int n) {
    sumx[0]=sumy[0]=0;
    ssumx[0]=ssumy[0]=0;
    for (int i=1;i<=n;i++) {
        scanf("%lld%lld", &x[i], &y[i]);
        sumx[i]=sumx[i-1]+x[i];sumy[i]=sumy[i-1]+y[i];
    }
    for (int i=1;i<=n;i++) {
        ssumx[i]=ssumx[i-1]+sumx[i];
        ssumy[i]=ssumy[i-1]+sumy[i];
    }
}
ll cala(int rev,int l,int r) {
    ll ret;
    int mid=(l+r)>>1;
    if (rev) {
        ret=2*(ssumy[r]-ssumy[mid]-(ll)(r-mid)*sumy[mid])-(ll)(r-l+2)*(sumy[r]-sumy[mid]);
    } else {
        ret=2*(ssumx[r]-ssumx[mid]-(ll)(r-mid)*sumx[mid])-(ll)(r-l+2)*(sumx[r]-sumx[mid]);
    }
    return ret;
}
ll solve(int n) {
    int i,j,k,now,pre;
    ll ret=INF;now=pre=1;
    dp[now][0][0]=dp[now][0][1]=INF;
    for (i=2;i<=n;i++) {
        pre=now;now^=1;
        for (j=0;j


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