zoj 3209 Treasure Map

来源：互联网发布：linux export 编辑：程序博客网时间：2024/05/20 18:51

Problem

acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3209

vjudge.net/contest/65998#problem/B

Meaning

要在 p 张矩形藏宝图碎片中选若干张，恰好铺满一张 n * m 的格子图，问最少需要几张。

Analysis

就是选若干张藏宝图，使得每个格子都有且只有被覆盖一次，就是精确覆盖问题：藏宝图当成行，每一个格子都当成一列，藏宝图能覆盖的格子就是 1，不能的就是 0。

相当于把二维的格子图压成一维的，link 的时候要算出每个格子位于哪一列，就要对格子按某种顺序排序。比如：

然后观察下坐标和格子编号的关系就知道怎么 link 了。

Source code

#include <cstdio>#include <algorithm>using namespace std;const int N = 30, M = 30, P = 500, NUM = P*N*M + N*M + 1;int up[NUM], down[NUM], left[NUM], right[NUM];int row[NUM], col[NUM], head[P+1], sum[N*M+1];int ans;void link(int r, int c, int id){row[id] = r;col[id] = c;up[id] = up[c];down[id] = c;up[c] = down[up[c]] = id;++sum[c];if(head[r] == -1)head[r] = left[id] = right[id] = id;else{left[id] = left[head[r]];right[id] = head[r];left[head[r]] = right[left[head[r]]] = id;}}void remove(int c){left[right[c]] = left[c];right[left[c]] = right[c];for(int i=down[c]; i!=c; i=down[i])for(int j=right[i]; j!=i; j=right[j]){up[down[j]] = up[j];down[up[j]] = down[j];--sum[col[j]];}}void resume(int c){left[right[c]] = right[left[c]] = c;for(int i=up[c]; i!=c; i=up[i])for(int j=left[i]; j!=i; j=left[j]){up[down[j]] = down[up[j]] = j;++sum[col[j]];}}int dance(int cnt){if(!right[0])return cnt;int c = right[0];for(int i=right[c]; i; i=right[i])if(sum[i] < sum[c])c = i;remove(c);for(int i=down[c]; i!=c; i=down[i]){for(int j=right[i]; j!=i; j=right[j])remove(col[j]);if(int tmp = dance(cnt+1))ans = min(ans, tmp); // 不直接return，继续求for(int j=left[i]; j!=i; j=left[j])resume(col[j]);}resume(c);return 0;}int main(){int t;scanf("%d", &t);while(t--){int n, m, p;scanf("%d%d%d", &n, &m, &p);for(int i=0; i<=n*m; ++i){up[i] = down[i] = col[i] = i;row[i] = sum[i] = 0;left[i] = i - 1;right[i] = i + 1;}left[0] = n * m;right[n*m] = 0;for(int i=1, id=n*m; i<=p; ++i){head[i] = -1;int x1, x2, y1, y2;scanf("%d%d%d%d", &x1, &y1, &x2, &y2);for(int j=x1; j<x2; ++j)for(int k=y1+1; k<=y2; ++k)link(i, j*m+k, ++id);}ans = p + 1;dance(0);if(ans <= p)printf("%d\n", ans);elseputs("-1");}return 0;}

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