标签：mem str cto space highlight als name back sign
题意：给定 n 个区间，让你进行合并，问你最后的区间是，如果是空集，输出 false 如果区间是是 [-32768，32767] ，则是true。
析：进行区间合并，要注意，如果是 x >= 0 && x <= 32767 那么输出是 x >= 0，在这地方，真是错死了。。。。。。后来看了数据才知道有这个，其他的就是进行区间合并，如果第 i 个和第 j 个区间合并时，假设 x >= a && x <= b || x >= c && x <= d，那么如果 b+1 >= c && a <= c || d+1 >= a && c <= a 就进行合并，当然对于单向的就进行一个标记左端点是-32768右端点是32767，也就是最后在输出。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e15;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1000 + 500;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Node{
int small, big;
bool flag;
};
Node a[maxn];
int clac(const string &s){
int ans = 0;
int sig = s[0] == ‘-‘ ? -1 : 1;
ans = s[0] == ‘-‘ ? 0 : s[0] - ‘0‘;
for(int i = 1; i < s.sz; ++i)
ans = ans * 10 + s[i] - ‘0‘;
return ans * sig;
}
vector<Node> ans;
int main(){
freopen("hard.in", "r", stdin);
freopen("hard.out", "w", stdout);
string s;
int idx = 0;
while(getline(cin, s)){
a[idx].flag = 0;
if(s[2] == ‘>‘){
stringstream ss(s);
string t;
ss >> t; ss >> t;
ss >> t;
a[idx].big = clac(t);
if((ss >> t)) {
if(t == "||") a[idx].small = 32767;
else{
ss >> t >> t;
ss >> t;
a[idx].small = clac(t);
}
}
else a[idx].small = 32767;
}
else{
stringstream ss(s);
string t;
ss >> t; ss >> t;
ss >> t;
a[idx].small = clac(t);
a[idx].big = -32768;
}
if(a[idx].big < -32768 || a[idx].small > 32767) a[idx].flag = 1;
++idx;
}
bool ok = false;
for(int i = 0; i < idx; ++i){
if(a[i].small >= a[i].big){
ok = true;
}
}
if(!ok){ puts("false"); return 0; }
for(int k = 0; k < 10; ++k)
for(int i = 0; i < idx; ++i){
if(a[i].small < a[i].big) a[i].flag = 1;
if(a[i].flag) continue;
for(int j = 0; j < idx; ++j){
if(a[j].small < a[j].big) a[j].flag = 1;
if(i == j) continue;
if(a[j].flag) continue;
if(a[i].big <= a[j].small+1 && a[i].small >= a[j].small || a[j].big <= a[i].small+1 && a[i].small <= a[j].small){
a[i].small = max(a[i].small, a[j].small);
a[i].big = min(a[i].big, a[j].big);
a[j].flag = 1;
}
}
}
for(int i = 0; i < idx; ++i)
if(!a[i].flag) ans.pb(a[i]);
if(ans[0].small == 32767 && ans[0].big == -32768) puts("true");
else {
for(int i = 0; i < ans.sz; ++i){
if(ans[i].big > -32768){
printf("x >= %d", ans[i].big);
if(ans[i].small < 32767) printf(" && x <= %d", ans[i].small);
}
else printf("x <= %d", ans[i].small);
if(i != ans.sz-1) printf(" ||");
puts("");
}
}
return 0;
}
Gym 101190H Hard Refactoring (模拟坑题)
标签：mem str cto space highlight als name back sign
原文地址：http://www.cnblogs.com/dwtfukgv/p/7450039.html
