poj_1691 Painting A Board（dfs+拓扑）

Painting A Board

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3797		Accepted: 1879

DescriptionThe CE digital company has built an Automatic Painting Machine (APM) to paint a flat board fully covered by adjacent non-overlapping rectangles of different sizes each with a predefined color.To color the board, the APM has access to a set of brushes. Each brush has a distinct color C. The APM picks one brush with color C and paints all possible rectangles having predefined color C with the following restrictions:To avoid leaking the paints and mixing colors, a rectangle can only be painted if all rectangles immediately above it have already been painted. For example rectangle labeled F in Figure 1 is painted only after rectangles C and D are painted. Note that eachrectangle must be painted at once, i.e. partial painting of one rectangle is not allowed.You are to write a program for APM to paint a given board so that the number of brush pick-ups is minimum. Notice that if one brush is picked up more than once, all pick-ups are counted.InputThe first line of the input file contains an integer M which is the number of test cases to solve (1 <= M <= 10). For each test case, the first line contains an integer N, the number of rectangles, followed by N lines describingthe rectangles. Each rectangle R is specified by 5 integers in one line: the y and x coordinates of the upper left corner of R, the y and x coordinates of the lower right corner of R, followed by the color-code of R.Note that:Color-code is an integer in the range of 1 .. 20.Upper left corner of the board coordinates is always (0,0).Coordinates are in the range of 0 .. 99.N is in the range of 1..15.OutputOne line for each test case showing the minimum number of brush pick-ups.Sample Input

1
7
0 0 2 2 1
0 2 1 6 2
2 0 4 2 1
1 2 4 4 2
1 4 3 6 1
4 0 6 4 1
3 4 6 6 2

Sample Output

3

将矩形设为顶点，若矩形i有紧贴着下边缘的矩形j，则i到j连一条有向边，这样很明显就是按拓扑序涂色了。

直接暴力就能过，想了一下，数据量不大，用一个整数来表示目前矩形涂色与否的状态，然后根据状态与目前刷子的颜色来记忆化搜索应该可行。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <bitset>#include <queue>#include <set>#include <map>#include <string>#include <algorithm>#define FOP freopen("data.txt","r",stdin)#define FOP2 freopen("data1.txt","w",stdout)#define inf 0x3f3f3f3f#define maxn 22#define mod 1000000007#define PI acos(-1.0)#define LL long longusing namespace std;struct Node{int x1, y1, x2, y2, color, indegree;bool flag;Node(int _x1, int _y1, int _x2, int _y2, int _color) : x1(_x1), y1(_y1), x2(_x2), y2(_y2), color(_color){indegree = 0, flag = 0;}};struct Edge{int from, to;Edge(int u, int v) : from(u), to(v) {}};int n, m;vector<int> G[maxn];vector<Edge> edges;int ans;vector<Node> nodes;void init(){ans = inf;for(int i = 1; i <= n; i++) G[i].clear();edges.clear();nodes.clear();}void AddEdge(int from, int to){edges.push_back(Edge(from, to));m = edges.size();G[from].push_back(m-1);}void dfs(int nn, int col, int cot){if(cot >= ans) return ; //剪枝if(nn == n){if(ans > cot) ans = cot;return ;}for(int i = 1; i <= n; i++){if(!nodes[i].flag && nodes[i].indegree == 0){nodes[i].flag = 1;for(int j = 0; j < G[i].size(); j++){Edge &e = edges[G[i][j]];nodes[e.to].indegree--;}if(nodes[i].color != col) dfs(nn+1, nodes[i].color, cot+1);else dfs(nn+1, col, cot);nodes[i].flag = 0;for(int j = 0; j < G[i].size(); j++){Edge &e = edges[G[i][j]];nodes[e.to].indegree++;}}}}int main(){int T;scanf("%d", &T);while(T--){init();scanf("%d", &n);int x1, x2, y1, y2, color;nodes.push_back(Node(0, 0, 0, 0, 0));for(int i = 1; i <= n; i++){scanf("%d%d%d%d%d", &y1, &x1, &y2, &x2, &color);nodes.push_back(Node(x1, y1, x2, y2, color));}for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){if(i == j) continue;if(nodes[i].y2 == nodes[j].y1 && nodes[i].x2 > nodes[j].x1 && nodes[i].x1 < nodes[j].x2){AddEdge(i, j);nodes[j].indegree++;}}}dfs(0, 0, 0);printf("%d\n", ans);}return 0;}