• cf1043E. Mysterious Crime(二分 前缀和)


    题意

    题目链接

    Sol

    考场上做完前四题的时候大概还剩半个小时吧,那时候已经困的不行了。

    看了看E发现好像很可做??

    又仔细看了几眼发现这不是sb题么。。。

    先考虑两个人,假设贡献分别为((x, y) (a, b))

    有两种组合方式,一种是(x + b),另一种是(y + a)

    (x + b >= y + a)

    那么(x - y >= a - b)

    因此我们按照(x - y)排序,对于每个位置,肯定是某一个前缀全选(x+b),除此之外都是(y+a)

    二分之后前缀和后缀和安排一下

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<map>
    #include<vector>
    #include<set>
    #include<queue>
    #include<cmath>
    //#include<ext/pb_ds/assoc_container.hpp>
    //#include<ext/pb_ds/hash_policy.hpp>
    #define Pair pair<int, int>
    #define MP(x, y) make_pair(x, y)
    #define fi first
    #define se second
    #define int long long
    #define LL long long
    #define ull unsigned long long
    #define rg register
    #define pt(x) printf("%d ", x);
    //#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
    //char buf[(1 << 22)], *p1 = buf, *p2 = buf;
    //char obuf[1<<24], *O = obuf;
    //void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
    //#define OS  *O++ = ' ';
    using namespace std;
    //using namespace __gnu_pbds;
    const int MAXN = 6e5 + 10, INF = 1e9 + 10, mod = 998244353;
    const int base = 137;
    const double eps = 1e-9;
    inline int read() {
        char c = getchar();
        int x = 0, f = 1;
        while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
        while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
        return x * f;
    }
    int N, M, x[MAXN], y[MAXN], cha[MAXN];
    struct Node {
        int x, y, val;
        bool operator < (const Node &rhs) const {
            return val < rhs.val;
        }
    }a[MAXN];
    int sx[MAXN], sy[MAXN], ans[MAXN];
    int calc(int x, int y, int a, int b) {
        return min(x + b, y + a);
    }
     main() {
        N = read(); M = read();
        for(int i = 1; i <= N; i++) {
            x[i] = a[i].x = read(), y[i] = a[i].y = read();
            cha[i] = a[i].val = a[i].x - a[i].y;
        }   
        for(int i = 1; i <= M; i++) {
            int p = read(), q = read();
            int mn = calc(a[p].x, a[p].y, a[q].x, a[q].y);
            ans[p] -= mn; ans[q] -= mn;
        }
        sort(a + 1, a + N + 1);
        sort(cha + 1, cha + N + 1);
        for(int i = 1; i <= N; i++) sx[i] = sx[i - 1] + a[i].x;
        for(int i = N; i >= 1; i--) sy[i] = sy[i + 1] + a[i].y;
        for(int i = 1; i <= N; i++) {
            int pos = upper_bound(cha + 1, cha + N + 1, x[i] - y[i]) - cha - 1;
            ans[i] += y[i] * pos + sx[pos] + (N - pos) * x[i] + sy[pos + 1] - calc(x[i], y[i], x[i], y[i]);
        }
        for(int i = 1; i <= N; i++) cout << ans[i] << " ";
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/9874646.html
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