[2016-03-16][HDU][3038][How Many Answers Are Wrong]

• 时间:2016-03-16 16:12:28 星期三

• 题目编号:[2016-03-16][HDU][3038][How Many Answers Are Wrong]

• 题目大意:

• 给定n个数字,给出若干个区间的和,问这些和有多少个是错误的,如果未知,那么就默认是对的,如果前面已知,就判断是否正确

• 分析:求出那些数据是已经给出的,已经给出的就可以求

• 方法:并查集更新已经出现的区间,集合内的元素为区间内的点,父节点表示区间的左端点,权值表示到父节点到当前点这段区间元素的和

• 解题过程遇到问题:多组数据!!!!

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
typedef long long LL;
#define CLR(x,y) memset((x),(y),sizeof((x)))
#define FOR(x,y,z) for(int (x)=(y);(x)<(z);++(x))
#define FORD(x,y,z) for(int (x)=(y);(x)>=(z);--(x))
const int maxn = 200000 + 100;
int fa[maxn],sum[maxn];
void ini(int n){
        CLR(sum,0);
        for(int i = 0; i <= n;++i){
                fa[i] = i;
        }
}
int fnd(int x){
        if(x == fa[x])  return x;
        int fax = fa[x];
        fa[x] = fnd(fa[x]);
        sum[x] += sum[fax];
        return fa[x];         
}
int main(){
        //freopen("in.txt","r",stdin);
        //freopen("out.txt","w",stdout);
        int n,m;             
        while(~scanf("%d%d",&n,&m)){
                int u,v,tsum,ans = 0;
                ini(n);
                FOR(i,0,m){               
                        scanf("%d%d%d",&u,&v,&tsum);
                        int t1,t2;
                        --u;
                        t1 = fnd(u);
                        t2 = fnd(v);
                        if(t1 == t2){                               
                                ans += (sum[v] - sum[u] != tsum);
                        }else {
                                fa[t2] = t1;
                                sum[t2] = tsum + sum[u] - sum[v];
                        }
                }
                printf("%d\n",ans);
        }
        return 0;
}

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