題面

預計得分：70 + 60 + 30 = 160

實際得分：40 + 60 + 0 = 100

T1數組開小了

T2比賽結束後5min AC

T3加了個記憶話搜索wa了、、

T1

zbq吊打std啊Orz

此題$O(nlog)$做法：

一個很顯然的思路：對每個做括號維護一個大根堆，每次取最大的。

但是這樣有不優的情況，比如$()), 1, 3, 5$

那麽我們還需要對每個已經加入的右括號維護一個小根堆。每次判斷是否替換掉更小的會更優

#include<cstdio>
#include<algorithm>
#include<queue>
#include
 
<vector>
#define LL long long 
using namespace std;
const int MAXN = 2 * 1e5 + 10, INF = 1e9;
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;
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 + 1);
    for(int i = 1; i <= N; i++) a[i] = read();
    int ans = 0;
    for(int i = 1; i <= N; i++) {
        if(s[i] == ‘(‘) mx.push(a[i]);
        if(s[i] == ‘)‘)
            if(!mx.empty() && a[i] + mx.top() > 0) {
                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 0;
}

T2

很顯然每個位置就那麽幾種可能

直接暴力判斷就好，前綴和優化

/*
60：直接BFS 
*/
#include<cstdio>
#include<algorithm>
#include<queue>
#define LL long long 
using namespace std;
const int MAXN = 1e5 + 10, INF = 1e9;
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, K;
int a[1001][1001], down[MAXN], vis[1001][1001];
struct Node {
    int xx1, yy1, xx2, yy2;
}p[MAXN];
bool pd(int x, int y) {
    if(x < 1 || x > N || y < 1 || y > M) return 1;
    return 0;
}
int hsum[1001][1001], lsum[1001][1001]; 
bool line(int x1, int y11, int x2, int y2, int id) {
    if(pd(x1, y11) || pd(x2, y2)) return 1;
    if(x1 == x2) {
        if(y11 > y2) swap(y11, y2);
        if(hsum[x1][y2] - hsum[x1][y11 - 1] == 0) return 0;
        if(a[x1][y11] == id && hsum[x1][y2] - hsum[x1][y11] == 0) return 0;
        if(a[x1][y2] == id && hsum[x1][y2 - 1] - hsum[x1][y11 - 1] == 0) return 0;  
        return 1;
    }
    if(y11 == y2) {
        if(x1 > x2) swap(x1, x2);
        if(lsum[x2][y2] - lsum[x1 - 1][y2] == 0) return 0;
        if(a[x1][y11] == id && lsum[x2][y2] - lsum[x1][y2] == 0) return 0;
        if(a[x2][y11] == id && lsum[x2 - 1][y2] - lsum[x1 - 1][y2] ==0) return 0;
        return 1;
    }
}
int main() {
    freopen("linking.in", "r", stdin);
    freopen("linking.out", "w", stdout);
    N = read(); M = read(); K = read();
    for(int i = 1; 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 = 1; i <= N; i++) 
        for(int j = 1; j <= M; j++)
            hsum[i][j] = hsum[i][j - 1] + a[i][j], 
            lsum[i][j] = lsum[i - 1][j] + a[i][j];
    int ans = 0;
    for(int i = 1; i <= K; i++) {
        int xx1 = p[i].xx1, yy1 = p[i].yy1, xx2 = p[i].xx2, yy2 = p[i].yy2, flag = 0;
        if(yy1 > yy2) swap(yy1, yy2), swap(xx1, xx2);
        for(int k = 1; k <= N; k++) 
            if(!line(xx2, yy2, k, yy2, i) && !line(xx1, yy1, k, yy1, i) && !line(k, yy1, k, yy2, i))
                {ans++; flag = 1; break;}
        if(flag == 1) continue;    
        for(int k = 1; 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 0;
}
/*
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

神仙題。

很顯然答案是一棵樹，那麽直接書上倍增就好

滿分做法不會。。

2018.7.21NOIP模擬賽？解題報告