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The problem is to give you a n*n matrix, there are k planets, we can let the weapon attack matrix of a row or a column to make the planet is crushed, now ask you at least a few of these weapons to smash all the planets, first we can know each weapon up to attack one line, so the maximum number of n Weapons, in addition we can view the weapon as a vertex, the planet as the edge, connect a map, find out the minimum vertex coverage of the graph is the maximum number of matches, the code is as follo

Click to open linkTest instructions: Place a bomb at a point where the bomb can be blown to infinity and is the form of a cross, but if it blows to # or * then stops in this direction, ask how many bombs are needed at least and detonate to blow up all the #Idea: A template dance chain, for the building we will all the # as a column, and then all the point is the place to put the bomb as a row, and then find each point can be fried # on it, after the building to run a dance chain can be repeated

#include HDOJ1150 (minimum point set overlay)

cases. And then followed T cases.The first line for each case contains integers n, m (0 Outputthe output should contain T lines. For each test case, you should just output a integer which is the least number of states the king has to divide into.Sample Input13 21 21 3 Sample Output2Source2011 multi-university Training Contest 3-host by BITRecommendlcyTest instructions: N Cities M-bar has an edge, divides these cities into several states, the principle is (1) you and V can reach each other if bo

Test instructions: All the roads of a castle form an n-node tree, and if a soldier is placed on a node, the side that is attached to the node is guarded,Ask for the minimum number of soldiers to keep all the sides guarded.Ideas: And the maximum independent set of ideas, similar to the transfer equation, with the 0,1 to indicate the root of the sub-tree not put soldiersDp[root][0] + = dp[son][1];Dp[root][1] + = min (dp[son][1],dp[son][0]);#include POJ 1463 Strategic Game (point set

Overwrite (override) Overloading (overload) Overwrite a method and rewrite it to achieve a different effect. Implementation of the interface method, inheriting the method of the child class overriding the parent class You can define some methods with the same name, but the input parameters are different Inheriting the method of the child class overriding the parent class1. The method's logo is exactly matched;2. The return value must be consistent;3. The exc

Topic Link: Click to open the linkThe main topic: given the coordinates of n points, now requires a square, completely surrounded by n points, and the smallest square area, the smallest square area.The expression cannot understand why the area changes with the angle is a single-peak function, waiting for Daniel to tell you that,If the area changes with the angle is a single-peak function, then naturally you can think of three points, according to the question asked to find the smallest square ar

= Head[u]; Head[u] = tot++;}voidInit () {scanf("%d%d", n, m);memset(Head,-1,sizeof(head)); tot =0; for(inti =1; I scanf("%d%d", node[i].x, AMP;NODE[I].Y); Addedge (node[i].x, NODE[I].Y); }}voidDfsintu) {pre[u] = lowlink[u] = ++dfs_clock;Stack[++top] = u; for(inti = Head[u]; I! =-1; i = e[i].next) {intv = e[i].to;if(!pre[v]) {DFS (v); Lowlink[u] = min (Lowlink[v], lowlink[u]); }Else if(!sccno[v]) {Lowlink[u] = min (Lowlink[u], pre[v]); } }intXif(Lowlink[u] = = Pre[

Drawable[] layers = new drawable[2]; Layers[0] = new colordrawable (primarycolor); LAYERS[1] = new colordrawable (0x7f000000); Layerdrawable layerdrawable = new layerdrawable (layers); Layerdrawable.setlayerinset (0, 0, 0, 0, 0); Layerdrawable.setlayerinset (1, 0, 0, 0, 0); SetBackground (layerdrawable);can also be layer-list used in XML to implementCopyright NOTICE: This article for Bo Master original article, without Bo Master permission not reprodu

,-1,0,0,-1};intMain () {intt,cas=1;scanf("%d", t); while(t--) {intNscanf("%d", n); for(intI=0; iscanf('%s ', S[i]);//figure to read all, and then to build the diagram} for(intI=0; i//Set up a two-part diagram for(intj=0; jif((i+j)%2==0s[i][j]==' # '){ for(intk=0;k4; k++) {intxx=dir[k][0]+i;intyy=dir[k][1]+j;if(xx>=0xx0yy' # ') {G[I*N+J].PB (XX*N+YY); }}}} Nx=n*n;printf("Case%d:%d\n", Cas++,hungar ()); for(intI=0; ireturn 0;} Copyright NOTICE: This article for Bo Master original artic

You can reduce the request time and make it more efficient by placing the required graphics on one sheet, partially obscuring them, and leaving the desired graphs.The original image is as follows:The code is as follows:After running the code as follows:The result of this is that each graph is shown on top, and all the upper half is obscured, and if you want to move the diagram to a different location, you can use the div to set it up.Multi-Graphic Overlay