UVALive 2038 Strategic game (树形DP,4级)
2013-08-28 16:18
543 查看
L - Strategic gameCrawling in process...Crawling failedTime Limit:3000MSMemory Limit:0KB 64bit IO Format:%lld & %lluSubmitStatusAppoint description:System Crawler (2013-06-01)Description
Bob enjoys playing computer games, especially strategic games, but sometimes he cannot find the solution fast enough and then he is very sad. Now he has the following problem. He must defend a medieval city, the roads of which form a tree. He has to putthe minimum number of soldiers on the nodes so that they can observe all the edges. Can you help him?Your program should find the minimum number of soldiers that Bob has to put for a given tree.For example for the tree:
the solution is one soldier (at the node 1).InputThe input file contains several data sets in text format. Each data set represents a tree with the following description:
![](https://icpcarchive.ecs.baylor.edu/components/com_onlinejudge/images/button_pdf.png)
![](https://icpcarchive.ecs.baylor.edu/external/20/p2038a.gif)
the number of nodesthe description of each node in the following format:node_identifier:(number_of_roads) node_identifier1 node_identifier2
node_identifiernumber_of_roadsornode_identifier:(0)
The node identifiers are integer numbers between 0 and n-1, forn nodes (0 < n ≤ 1500). Every edge appears only once in the input data.OutputThe output should be printed on the standard output. For each given input data set, print one integer number in a single line that gives the result (the minimum number of soldiers).Sample Input4 0:(1) 1 1:(2) 2 3 2:(0) 3:(0) 5 3:(3) 1 4 2 1:(1) 0 2:(0) 0:(0) 4:(0)Sample Output
1 2
思路:简单树形DP dp[u][0]不选dp[u][1]选。
dp[u][1]+=min(dp[v][0],dp[v][1])
dp[u][0]+=dp[v][1];
数据生成器
#include<iostream> #include<cstdio> #include<cstring> #include<ctime> #include<cstdlib> using namespace std; bool vis[1600][1600]; int main() { int n; memset(vis,0,sizeof(vis)); for(int i=0;i<=1503;++i)vis[i][i]=1; srand((int)time(0)); n=rand()%1500+1;printf("%d\n",n); int z=n; for(int i=0;i<n;++i) { int zz=rand()%100; if(z>zz) z-=zz; else zz=0; printf("%d:(%d)",i,zz); int k=0; for(int j=0;j<zz;++j) { while(k<n&&vis[i][k])++k; printf(" %d",k);vis[i][k]=vis[k][i]=1; } printf("\n"); } }对拍程序[/code]
@echo off:looprand.exe>data.txtstd.exe<data.txt>std.txtmy.exe<data.txt>my.txtfc my.txt std.txtif not errorlevel 1 goto looppausegoto loop代码[/code]
#include<cstring>#include<iostream>#include<cstdio>#define FOR(i,a,b) for(int i=a;i<=b;++i)#define clr(f,z) memset(f,z,sizeof(f))using namespace std;const int mm=1502;const int oo=0x3f3f3f3f;int head[mm],edge;int dp[mm][2];class Edge{public:int v,next;}e[mm*mm];void data(){edge=0;clr(head,-1);}void add(int u,int v){e[edge].v=v;e[edge].next=head[u];head[u]=edge++;}int DP(int u,int fa,int yes){ int v;if(dp[u][yes]!=-1)return dp[u][yes];dp[u][yes]=yes;for(int i=head[u];~i;i=e[i].next){v=e[i].v;if(v==fa)continue;if(yes)dp[u][yes]+=min(DP(v,u,0),DP(v,u,1));else dp[u][yes]+=DP(v,u,1);}return dp[u][yes];}int main(){int n,a,b,m;while(~scanf("%d",&n)){data();FOR(i,1,n){scanf("%d",&a);scanf(":(%d)",&m);FOR(j,1,m){scanf("%d",&b);add(a,b);add(b,a);//cout<<a<<" "<<b<<endl;}}clr(dp,-1);int ans=min(DP(0,-1,0),DP(0,-1,1));printf("%d\n",ans);}}[/code]
相关文章推荐
- UVALive 2038 Strategic game (树形DP,4级)
- 【树形dp】UVALive 2038 Strategic game
- UVALive 2038 - Strategic game (经典树形DP)
- UVALive 2038 Strategic game--树形dp
- UVa 2038 - Strategic game(二分图最小顶点覆盖 or 树形DP)
- UVALive2038(树形dp)
- UVA1292-----Strategic game-----树形DP解决树上的最小点覆盖问题
- uva 1292 - Strategic game(树形dp)
- uvalive 5088 hdu3066(树形dp)
- UVALive 4256 Salesmen (树DP,4级)
- UVALIVE 3346 Perfect Domination on Trees 树形DP
- UVALive 4015 - Caves(树形DP)
- UVA 1292 - Strategic game(树形dp)
- UVA1292-----Strategic game-----树形DP解决树上的最小点覆盖问题
- UVALive 5088 Alice and Bob's Trip(树形DP)
- UVALive 3412 Pesky Heroes(树形dp)
- LA 2038 Strategic game(最小点覆盖,树形dp,二分匹配)
- UVA Live Archive 4015 Cave (树形dp,分组背包)
- Uva 1292 - Strategic game 树形dp 最小点覆盖
- UVALive 4015 树形dp