http://acm.hdu.edu.cn/showproblem.php?pid=3308
又是一题区间合并的题，今晚就这题就debug到我晕了。。。。。
同样是之前poj那题的错误，就是区间合并的部分思路混乱了一下。
题目很简洁，就是要求区间的最大上升子串，同时要有动态的单点更新！线段树功能【单点更新，查询区间】。
题目告诉我，我应该找多点这样的题来做，从而防止这样的思维混乱再次发生，因为这明显就是应该1y的水题！- -
View Code
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <algorithm> 5 #include <vector> 6 #include <cassert> 7 8 using namespace std; 9 10 #define lson l, m, rt << 1 11 #define rson m + 1, r, rt << 1 | 1 12 const int maxn = 100005; 13 14 int lval[maxn << 2], rval[maxn << 2], llen[maxn << 2], rlen[maxn << 2], mx[maxn << 2]; 15 struct Node{ 16 int lval, rval; 17 int llen, rlen; 18 int mx; 19 Node() {} 20 Node(int lv, int rv, int ll, int rl, int m): 21 lval(lv), rval(rv), llen(ll), rlen(rl), mx(m) {} 22 }; 23 24 void up(int rt, int len){ // 关键位置，容易发生错漏 25 int l = rt << 1; 26 int r = rt << 1 | 1; 27 int m1 = len >> 1, m2 = len - m1; 28 29 mx[rt] = max(mx[l], mx[r]); 30 llen[rt] = llen[l]; 31 rlen[rt] = rlen[r]; 32 if (rval[l] < lval[r]){ 33 mx[rt] = max(mx[rt], rlen[l] + llen[r]); 34 if (llen[l] == m2) llen[rt] = llen[l] + llen[r]; 35 if (rlen[r] == m1) rlen[rt] = rlen[r] + rlen[l]; 36 } 37 lval[rt] = lval[l]; 38 rval[rt] = rval[r]; 39 } 40 41 void build(int l, int r, int rt){ 42 if (l == r){ 43 int tmp; 44 45 scanf("%d", &tmp); 46 lval[rt] = rval[rt] = tmp; 47 mx[rt] = llen[rt] = rlen[rt] = 1; 48 49 return ; 50 } 51 int m = (l + r) >> 1; 52 53 build(lson); 54 build(rson); 55 up(rt, r - l + 1); 56 //printf("%d %d : len %d %d val %d %d mx %d\n", l, r, llen[rt], rlen[rt], lval[rt], rval[rt], mx[rt]); 57 } 58 59 void modify(int pos, int key, int l, int r, int rt){ 60 if (l == r){ 61 //assert(l == pos); 62 lval[rt] = rval[rt] = key; 63 64 return ; 65 } 66 int m = (l + r) >> 1; 67 68 if (pos <= m) modify(pos, key, lson); 69 else modify(pos, key, rson); 70 up(rt, r - l + 1); 71 } 72 73 Node query(int L, int R, int l, int r, int rt){ 74 Node ret; 75 if (L <= l && r <= R){ 76 ret = Node(lval[rt], rval[rt], llen[rt], rlen[rt], mx[rt]); 77 78 return ret; 79 } 80 int m = (l + r) >> 1; 81 Node tmp; 82 // 下面的也是关键位置，合并绝对不能水 83 if (L <= m){ 84 int m1 = min(m - L + 1, m - l + 1); 85 int m2 = min(R - m, r - m); 86 87 ret = query(L, R, lson); 88 //printf("%d %d : mx %d len %d %d val %d %d\n", l, m, ret.mx, ret.llen, ret.rlen, ret.lval, ret.rval); 89 if (m < R){ 90 tmp = query(L, R, rson); 91 //printf("%d %d : mx %d len %d %d val %d %d\n", m + 1, r, tmp.mx, tmp.llen, tmp.rlen, tmp.lval, tmp.rval); 92 ret.mx = max(ret.mx, tmp.mx); 93 if (ret.rval < tmp.lval){ 94 ret.mx = max(ret.mx, ret.rlen + tmp.llen); 95 } 96 if (ret.llen == m1 && ret.rval < tmp.lval) ret.llen += tmp.llen; 97 if (tmp.rlen == m2 && ret.rval < tmp.lval) ret.rlen += tmp.rlen; 98 else ret.rlen = tmp.rlen; 99 ret.rval = tmp.rval; 100 } 101 return ret; 102 } 103 if (m < R) ret = query(L, R, rson); 104 //printf("%d %d : mx %d len %d %d val %d %d\n", m + 1, r, ret.mx, ret.llen, ret.rlen, ret.lval, ret.rval); 105 106 return ret; 107 } 108 109 void deal(int n, int m){ 110 char buf[3]; 111 112 build(0, n - 1, 1); 113 while (m--){ 114 int l, r; 115 116 scanf("%s%d%d", buf, &l, &r); 117 //assert(buf[0] == 'Q' || buf[0] == 'U'); 118 if (buf[0] == 'Q'){ 119 printf("%d\n", query(l, r, 0, n - 1, 1).mx); 120 } 121 else if (buf[0] == 'U'){ 122 modify(l, r, 0, n - 1, 1); 123 } 124 } 125 } 126 127 int main(){ 128 int n, m; 129 int T; 130 131 //freopen("in", "r", stdin); 132 scanf("%d", &T); 133 while (T-- && ~scanf("%d%d", &n, &m)){ 134 deal(n, m); 135 } 136 137 return 0; 138 }
——written by Lyon