NOIP2011 Day2 T2 聪明的质监员

来源：互联网发布：调音师知乎编辑：程序博客网时间：2024/05/17 23:45

W越大，Y越小，所以可以二分W使Y接近S。
少用min、abs可以让思路更清晰。

#include<cstdio>#include<cstring>#include<cmath>#include<iostream>using namespace std;const int maxn = 200010;int n, m, num[maxn];int mw;long long s, sum[maxn], tot;struct node1{    int w, v;} a[maxn];struct node2{    int li, ri;} b[maxn];long long judge(int x){    memset(sum, 0, sizeof(sum));    memset(num, 0, sizeof(num));    for(int i = 1; i <= n; i++){        sum[i] = sum[i-1], num[i] = num[i-1];        if(a[i].w > x){            sum[i] += a[i].v;            num[i]++;        }    }    tot = 0;    for(int i = 1; i <= m; i++)        tot += (sum[b[i].ri] - sum[b[i].li-1])*(num[b[i].ri] - num[b[i].li-1]);    return tot;}int main(){    cin >> n >> m >> s;    for(int i = 1; i <= n; i++){        cin >> a[i].w >> a[i].v;        mw = max(mw, a[i].w);    }    for(int i = 1; i <= m; i++)        cin >> b[i].li >> b[i].ri;    long long ans = s;    int l = 0, r = mw + 1;    while(l < r){        int mid = (l+r) >> 1;        long long now = judge(mid);        if(now > s){            l = mid + 1;            if(ans > now - s)   ans = now - s;        }        else if(now < s){            r = mid;            if(ans > s - now)   ans = s - now;        }        else{            cout << "0" << endl;            return 0;        }    }    cout << ans << endl;    return 0;}

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