HDU - 1677Nested Dolls最長上升子序列變式

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HDU - 1677Nested Dolls Time Limit: 1000MS   Memory Limit: 32768KB   64bit IO Format: %I64d & %I64u Submit Status Description Dilworth is the world’s most prominent collector of Russian nested dolls: he literally has thousands of them! You know, the wooden hollow dolls of different sizes of which the smallest doll is contained in the second smallest, and this doll is in turn contained in the next one and so forth. One day he wonders if there is another way of nesting them so he will end up with fewer nested dolls? After all, that would make his collection even more magnificent! He unpacks each nested doll and measures the width and height of each contained doll. A doll with width w1 and height h1 will fit in another doll of width w2 and height h2 if and only if w1 < w2 and h1 < h2. Can you help him calculate the smallest number of nested dolls possible to assemble from his massive list of measurements?    Input On the first line of input is a single positive integer 1 <= t <= 20 specifying the number of test cases to follow. Each test case begins with a positive integer 1 <= m <= 20000 on a line of itself telling the number of dolls in the test case. Next follow 2m positive integers w1, h1,w2, h2, . . . ,wm, hm, where wi is the width and hi is the height of doll number i. 1 <= wi, hi <= 10000 for all i.    Output For each test case there should be one line of output containing the minimum number of nested dolls possible.    Sample Input 4320 30 40 50 30 40420 30 10 10 30 20 40 50310 30 20 20 30 10410 10 20 30 40 50 39 51   Sample Output 1232 大概意思與最小攔截系統類別似  第一種方法，自己寫二分函數 /*Problem : 1677 ( Nested Dolls ) Judge Status : AcceptedRunId : 14351565 Language : G++ Author : 24862486*/#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>using namespace std;const int maxn = 20000 + 5;const int INF=0x3f3f3f3f;struct point { int h, w; point(int w ,int h) : w(w), h(h) {} point() {}} ps[maxn];int dp[maxn], t, m;bool cmp(const point &a,const point &b) { if(a.w == b.w) return a.h < b.h; return a.w > b.w;}int solve() { memset(dp,0,sizeof(dp)); int top = 0, maxx = -1; dp[top] = -1; for(int i = 1; i <= m; i ++) { if(ps[i].h >= dp[top]) { dp[++ top] = ps[i].h;//其中dp[1]--dp[top]一定是不下降子序列，因為每一次的新系統的產生，表示大於他的值 } else { int l = 0,r = top;//(0,top] while(r - l > 1) { int mid = (l + r) >> 1; if(ps[i].h < dp[mid]) {//是不能夠相等，相等則依舊取不到 r = mid; } else l = mid; } dp[r] = ps[i].h; } } return top;}int main() { //freopen("D://imput.txt", "r", stdin); scanf("%d", &t); while(t --) { scanf("%d", &m); for(int i = 1; i <= m; i ++) { scanf("%d%d", &ps[i].w, &ps[i].h); } sort(ps + 1,ps + m + 1,cmp); printf("%d\n",solve()); } return 0;} 第二種，運用upper_bound以及lower_bound函數 /*Problem : 1677 ( Nested Dolls ) Judge Status : AcceptedRunId : 14351713 Language : G++ Author : 24862486*/#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>using namespace std;const int maxn = 20000 + 5;const int INF=0x3f3f3f3f;struct point { int h, w; point(int w ,int h) : w(w), h(h) {} point() {} bool operator < (const point &object)const { return h < object.h; }} ps[maxn],dp[maxn];bool cmp(const point &a,const point &b) { if(a.w == b.w) return a.h < b.h; return a.w > b.w;}int main() { //freopen("D://imput.txt", "r", stdin); int t, m; scanf("%d", &t); while(t --) { scanf("%d", &m); for(int i = 0; i < m; i ++) { scanf("%d%d", &ps[i].w, &ps[i].h); } sort(ps,ps + m,cmp); for(int i = 0; i < m ; i ++) { dp[i].h = INF; dp[i].w = INF; } for(int i = 0; i < m ; i ++) { *upper_bound(dp,dp + m, ps[i]) = ps[i];//不是lower_bound()原因是不能夠相等 } printf("%d\n", lower_bound(dp, dp + m, point(INF,INF)) - dp); } return 0;}

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HDU - 1677Nested Dolls最長上升子序列變式