2018.7.21NOIP模拟赛？解题报告

题面

预计得分：70 + 60 + 30 = 160

实际得分：40 + 60 + 0 = 100

T1数组开小了

T2比赛结束后5min AC

T3加了个记忆话搜索wa了、、

T1

zbq吊打std啊Orz

此题$O(nlog)$做法：

一个很显然的思路：对每个做括号维护一个大根堆，每次取最大的。

但是这样有不优的情况，比如$()), 1, 3, 5$

那么我们还需要对每个已经加入的右括号维护一个小根堆。每次判断是否替换掉更小的会更优

#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
#define LL long long
using namespace std;
const int MAXN =  * 1e5 + , INF = 1e9;
inline int read() {
    char c = getchar(); int x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
int N;
priority_queue<int> mx;
priority_queue<int, vector<int>, greater<int> >mi;
char s[MAXN];
int a[MAXN];
int main() {
    freopen("bracket.in", "r", stdin);
    freopen("bracket.out", "w", stdout);
    N = read();
    scanf("%s", s + );
    for(int i = ; i <= N; i++) a[i] = read();
    int ans = ;
    for(int i = ; i <= N; i++) {
        if(s[i] == '(') mx.push(a[i]);
        if(s[i] == ')')
            if(!mx.empty() && a[i] + mx.top() > ) {
                if(mi.empty() || (!mi.empty() && mx.top() > - mi.top())) ans += a[i] + mx.top(), mx.pop(), mi.push(a[i]);
                else if(!mi.empty() && a[i] > mi.top()) ans -= mi.top(), mi.pop(), ans += a[i], mi.push(a[i]);
            } else if(!mi.empty() && a[i] > mi.top())  ans -= mi.top(), mi.pop(), ans += a[i], mi.push(a[i]);

    }
    printf("%d", ans);
    return ;
}

T2

很显然每个位置就那么几种可能

直接暴力判断就好，前缀和优化

/*
60：直接BFS
*/
#include<cstdio>
#include<algorithm>
#include<queue>
#define LL long long
using namespace std;
const int MAXN = 1e5 + , INF = 1e9;
inline int read() {
    char c = getchar(); int x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
int N, M, K;
int a[][], down[MAXN], vis[][];
struct Node {
    int xx1, yy1, xx2, yy2;
}p[MAXN];
bool pd(int x, int y) {
    if(x <  || x > N || y <  || y > M) return ;
    return ;
}
int hsum[][], lsum[][];
bool line(int x1, int y11, int x2, int y2, int id) {
    if(pd(x1, y11) || pd(x2, y2)) return ;
    if(x1 == x2) {
        if(y11 > y2) swap(y11, y2);
        if(hsum[x1][y2] - hsum[x1][y11 - ] == ) return ;
        if(a[x1][y11] == id && hsum[x1][y2] - hsum[x1][y11] == ) return ;
        if(a[x1][y2] == id && hsum[x1][y2 - ] - hsum[x1][y11 - ] == ) return ;
        return ;
    }
    if(y11 == y2) {
        if(x1 > x2) swap(x1, x2);
        if(lsum[x2][y2] - lsum[x1 - ][y2] == ) return ;
        if(a[x1][y11] == id && lsum[x2][y2] - lsum[x1][y2] == ) return ;
        if(a[x2][y11] == id && lsum[x2 - ][y2] - lsum[x1 - ][y2] ==) return ;
        return ;
    }
}
int main() {
    freopen("linking.in", "r", stdin);
    freopen("linking.out", "w", stdout);
    N = read(); M = read(); K = read();
    for(int i = ; i <= K; i++) {
        int xx1 = read(), yy1 = read(), xx2 = read(), yy2 = read();
        a[xx1][yy1] = i;
        a[xx2][yy2] = i;
        p[i] = (Node) {xx1, yy1, xx2, yy2};
    }
    for(int i = ; i <= N; i++)
        for(int j = ; j <= M; j++)
            hsum[i][j] = hsum[i][j - ] + a[i][j],
            lsum[i][j] = lsum[i - ][j] + a[i][j];
    int ans = ;
    for(int i = ; i <= K; i++) {
        int xx1 = p[i].xx1, yy1 = p[i].yy1, xx2 = p[i].xx2, yy2 = p[i].yy2, flag = ;
        if(yy1 > yy2) swap(yy1, yy2), swap(xx1, xx2);
        for(int k = ; k <= N; k++)
            if(!line(xx2, yy2, k, yy2, i) && !line(xx1, yy1, k, yy1, i) && !line(k, yy1, k, yy2, i))
                {ans++; flag = ; break;}
        if(flag == ) continue;
        for(int k = ; k <= M; k++)
            if(!line(xx2, yy2, xx2, k, i) && !line(xx2, k, xx1, k, i) && !line(xx1, yy1, xx1, k, i))
                {ans++; break;}
    }
    printf("%d", ans);
    return ;
}
/*
20 20 3
1 1 20 20
2 1 2 20
3 1 1 20

3 3 3
1 3 2 2
1 1 3 3
1 2 2 1

*/

T3

神仙题。

很显然答案是一棵树，那么直接书上倍增就好

满分做法不会。。