Strategic Game

　　　　　　　　　　　　Time Limit: 20000/10000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
　　　　　　　　　　　　　　　　　　Total Submission(s): 8504????Accepted Submission(s): 4094

Problem 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 put the 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.

The input file contains several data sets in text format. Each data set represents a tree with the following description:

the number of nodes
the description of each node in the following format
node_identifier:(number_of_roads) node_identifier1 node_identifier2 ... node_identifier
or
node_identifier:(0)

The node identifiers are integer numbers between 0 and n-1, for n nodes (0 < n <= 1500). Every edge appears only once in the input data.

For example for the tree:?

?

the solution is one soldier ( at the node 1).

The 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). An example is given in the following table: Sample Input 4 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 Source Southeastern Europe 2000 Recommend JGShining???|???We have carefully selected several similar problems for you:??1053?1151?1142?1233?1045?

問題描述
鮑勃喜歡玩電腦游戲，特別是戰略游戲，但有時候他無法快速找到解決方案，那么他很傷心。現在他有以下問題。他必須捍衛一座中世紀城市，其道路形成一棵樹。他必須將最少數量的士兵放在節點上，以便他們可以觀察所有的邊緣。你的程序應該找到Bob給給定樹的最小兵數。

思路：用最少的節點覆蓋所有的邊。最小點覆蓋裸題=最大匹配數

這個題的數據比較大，用鄰接表的話會T，用鄰街鏈表的話可以A

代碼：

#include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define N 1500+10 using namespace std; bool vis[N]; int n,m,k,x,y,tot,head[N],girl[N],map[N][N]; int read() {int x=0,f=1; char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1; ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}return x*f; } struct Edge {int from,to,next; }edge[N]; int add(int x,int y) {tot++;edge[tot].to=y;edge[tot].next=head[x];head[x]=tot; } int find(int x) {for(int i=head[x];i;i=edge[i].next){int t=edge[i].to;if(!vis[t]){vis[t]=true;if(girl[t]==-1||find(girl[t])) {girl[t]=x;return 1;}}}return 0; } int main() {int t=0,sum,ans;while(scanf("%d",&n)!=EOF){ans=0,sum=0;tot=0;memset(map,0,sizeof(map));memset(head,0,sizeof(head));for(int i=1;i<=n;i++){x=read();m=read();while(m--){y=read();add(x+1,y+1),add(y+1,x+1);}} memset(girl,-1,sizeof(girl));for(int i=1;i<=n;i++){memset(vis,0,sizeof(vis));if(find(i)) ans++;}printf("%d\n",ans/2);}return 0; }

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轉載于:https://www.cnblogs.com/z360/p/7435613.html

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