HDU4496-D-City

Problem Description

Luxer is a really bad guy. He destroys everything he met.
One day Luxer went to D-city. D-city has N D-points and M D-lines. Each D-line connects exactly two D-points. Luxer will destroy all the D-lines. The mayor of D-city wants to know how many connected blocks of D-city left after Luxer destroying the first K D-lines in the input.
Two points are in the same connected blocks if and only if they connect to each other directly or indirectly.

Input

First line of the input contains two integers N and M.
Then following M lines each containing 2 space-separated integers u and v, which denotes an D-line.
Constraints:
0 < N <= 10000
0 < M <= 100000
0 <= u, v < N.

Output

Output M lines, the ith line is the answer after deleting the first i edges in the input.

Sample Input

5 10 0 1 1 2 1 3 1 4 0 2 2 3 0 4 0 3 3 4 2 4

Sample Output

1 1 1 2 2 2 2 3 4 5

思路：题意就是一个人把一个连通的图，一条一条路慢慢地破坏，每破坏一条路，看是否有顶点被独立出来。我按照这个思路用一个数组储存了所有点的入度。最后遍历，每遍历一条路，路的端点减1；下面是这个思路的代码（不是超时就是wrong）

代码：

#include<iostream>
#include<queue>
#include<algorithm>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cstdio>
#include<cstdlib>
#include<cctype>
#include<cstring>
using namespace std;
const int MAX = 10010;
const int INF = 0X3f3f3f;

int indegree[MAX];


int main() {
	int n, m;
	while(scanf("%d%d", &n, &m) != EOF) {
        vector<pair<int, int> > t;
        int u, v;
        memset(indegree, 0, sizeof(indegree));
        int ans = 1;
        while(m--) {
            scanf("%d%d", &u, &v);
            t.push_back(make_pair(u,v));
            indegree[u]++;
            indegree[v]++;
        }
        for(int i = 0; i < t.size(); i++) {
            int u = t[i].first;
            int v = t[i].second;
            indegree[u]--;
            if(indegree[u] == 0) {
                ans++;
            }
            indegree[v]--;
            if(indegree[v] == 0) {
                ans++;
            }
            if(i == t.size() - 1)
                printf("%d
", ans - 1);
            else
                printf("%d
", ans);
        }
	}
	return 0;
}

后来看了题解。用的逆序并查集。先把所有点当作未连通的，每处理一条路（倒着处理，但是第一条《输入时候的第一条不作处理》），把连通块输出。

AC代码：

#include<iostream>
#include<queue>
#include<algorithm>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cstdio>
#include<cstdlib>
#include<cctype>
#include<cstring>
using namespace std;
const int MAX = 10010;
const int INF = 0X3f3f3f;

int father[MAX];
int n, m;
int sum;
void init() {//初始化
    for(int i = 0; i < n; i++) {
        father[i] = i;
    }
}
int findfather(int x) { //路径压缩
    int a = x;
    while(x != father[x]) {
        x = father[x];
    }
    while(a != father[a]) {
        int z = a;
        a = father[a];
        father[z] = x;
    }
    return x;
}

void Union(int a, int b) {//合并
    int faA = findfather(a);
    int faB = findfather(b);
    if(faA != faB) {
        sum--;//如果不是一个集合 就说明原来不连通的点 少了一个
        father[faA] = faB;
    }
}

int main() {
    int u, v;
    while(scanf("%d%d", &n, &m) != EOF) {
        init();
        sum = n;
        vector<pair<int, int> > t;//存路，以后遍历用
        vector<int> tt;//存输出数据
        tt.push_back(n);//先把最初的不连通  存进去；
        while(m--) {
            scanf("%d%d", &u, &v);
            t.push_back(make_pair(u, v));
        }
        for(int i = t.size() - 1; i > 0; i--) {//遍历路径  注意 i > 0, 而不是 i >= 0;
            int u = t[i].first;
            int v = t[i].second;
            Union(u, v);
            tt.push_back(sum);
        }
        for(int i = tt.size() - 1; i >= 0; i--) {//倒着输出
            printf("%d
", tt[i]);
        }
    }
	return 0;
}