ZOJ 3209 Treasure Map

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3209

dancing links 每个点作为一列。

每个块作为一行。

不知道为什么插节点的时候 如果把循环写成

for(int i = y1+1; i <= y2; i++){      //行 
              for(int j = x1+1; j <= x2; j++){ //列 
　　　　　　
              }
          }
然后col = (y-1)*n+x就会超时。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <iomanip>
using namespace std;
int T, n, m, p, x1, y1, x2, y2;
#define maxn 510*35*35
int R[maxn], L[maxn], U[maxn], D[maxn];
int H[maxn], C[maxn], colsum[999];
int head, cnt, ans;
void addnode(int i, int j, int h, int rowhead, int sum, int pre){
    int col = j+(i-1)*m; //cout<<col<<" ";
    cnt++;
    
    H[cnt] = h; C[cnt] = col;
    if(sum == 1){
        R[cnt] = cnt; L[cnt] = cnt;
    }
    else if(sum == 2){
        R[pre] = cnt; R[cnt] = pre;
        L[pre] = cnt; L[cnt] = pre; 
    }
    else{
        R[cnt] = rowhead; R[pre] = cnt;
        L[cnt] = pre; L[rowhead] = cnt;
    }
    
    D[U[col]] = cnt;
    U[cnt] = U[col];
    D[cnt] = col;
    U[col] = cnt;
    
    colsum[col]++; 
    
}
void init(){
    head = cnt = 0;
    ans = 510;  //最多就是500块。 
    R[head] = 1; L[head] = n*m; U[head] = head; D[head] = head;
    H[head] = 0; C[head] = 0;
    memset(colsum, 0, sizeof(colsum));
    for(int i = 1; i <= n*m; i++){
        cnt++;
        U[cnt] = D[cnt] = cnt;
        H[cnt] = 0; C[cnt] = i;
        if(i == n*m){
            R[cnt-1] = cnt; L[cnt] = cnt-1; R[cnt] = head;
        }    
        else{
            R[cnt-1] = cnt; L[cnt] = cnt-1;    
        }
    }
    for(int cc = 1; cc <= p; cc++){
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        int sum = 0, rowhead = -1, pre;
        for(int i = x1+1; i <= x2; i++){      //行 
            for(int j = y1+1; j <= y2; j++){ //列 
                sum++;
                addnode(i, j ,cc, rowhead, sum, pre);
                if(rowhead == -1) rowhead = cnt;
                pre = cnt;
            }
        }
    }
}
void remove(int c){
    R[L[c]] = R[c]; L[R[c]] = L[c];
    for(int i = D[c]; i != c; i = D[i]){
        for(int j = R[i]; j != i; j = R[j]){
            U[D[j]] = U[j]; D[U[j]] = D[j];
            colsum[C[j]]--;
        }
    }
}
void resume(int c){
    R[L[c]] = c; L[R[c]] = c;
    for(int i = D[c]; i != c; i = D[i]){
        for(int j = R[i]; j != i; j = R[j]){
            U[D[j]] = j; D[U[j]] = j;
            colsum[C[j]]++;
        }
    }
}
void dance(int k){
    if(k >= ans) return;
    int c = R[head];
    if(c == head){
        ans = k;
        return;
    }
    int min = 9999999;
    for(int i = R[head]; i != head; i = R[i]){
        if(colsum[i] <= min){
            min = colsum[i]; c = i;
        }
    }
    remove(c);
    for(int i = D[c]; i != c; i = D[i]){
        for(int j = R[i]; j != i; j = R[j]) remove(C[j]);
        dance(k+1);
        for(int j = L[i]; j != i; j = L[j]) resume(C[j]); //要写成L 
    }
    resume(c);
    return ;
}
int main(){
    scanf("%d", &T);
    while(T--){
        scanf("%d%d%d", &n, &m, &p); //n是列，m是行。 
        init();
        dance(0);
        if(ans == 510) printf("-1
");
        else printf("%d
", ans);
    }
    
    return 0;
}