題意：

給你一些聯(lián)通關(guān)系，問Bob先選擇一些路徑（1~n）聯(lián)通，Alice在路徑上染色，Bob的目的是選擇一些路徑使得染色變化最小，對于Alice來說，需要使得在Bob選擇的（$1?n$）d的路徑上使得顏色變化最大。

題目：

Alice and Bob are playing a game on a simple connected graph with N nodes and M edges.
Alice colors each edge in the graph red or blue.
A path is a sequence of edges where each pair of consecutive edges have a node in common. If the first edge in the pair is of a different color than the second edge, then that is a ‘‘color change.’’
After Alice colors the graph, Bob chooses a path that begins at node 1 and ends at node N. He can choose any path on the graph, but he wants to minimize the number of color changes in the path. Alice wants to choose an edge coloring to maximize the number of color changes Bob must make. What is the maximum number of color changes she can force Bob to make, regardless of which path he chooses? changes she can force Bob to make, regardless of which path he chooses?

輸入描述:

The first line contains two integer values N and M with $2\le N\le 100000$ and $1\le M\le 100000$. The next M lines contain two integers $a_i$and $b_i$ indicating an undirected edge between nodes $a_i$ and $b_i$ ($1\le a_i,b_i\le N,a_i=b_i)$All edges in the graph are unique.

輸出描述:

Output the maximum number of color changes Alice can force Bob to make on his route from node 1 to node N.

示例1

輸入

3 3
1 3
1 2
2 3

輸出

0

示例2

輸入

7 8
1 2
1 3
2 4
3 4
4 5
4 6
5 7
6 7

輸出

3

分析：

為了使得染色變化最小，那么就選擇1~n最短路徑即可，因為是無權(quán)路徑，且題目是求染色變化，可以用BFS在一個無權(quán)圖上求從起點(diǎn)到其他所有點(diǎn)的最短路徑。最大染色變化即為最短路徑長-1；

AC代碼：

#include<cstdio> #include<cstring> #include<algorithm> #include<vector> #include<queue> const int M=1e5+10; using namespace std; bool vis[M]; int n,m,dist[M]; vector<int>g[M]; void BFS() {queue<int>q;q.push(1);vis[1]=1;dist[1]=0;while(!q.empty()){int i=q.front();q.pop();for(int k=0; k<g[i].size(); k++){int j=g[i][k];if(vis[j])continue;vis[j]=1;dist[j]=dist[i]+1;q.push(j);}} } int main() {memset(vis,0,sizeof(vis));scanf("%d%d",&n,&m);for(int i=1; i<=m; i++){int x,y;scanf("%d%d",&x,&y);g[x].push_back(y);g[y].push_back(x);}BFS();printf("%d",dist[n]-1);return 0; }

總結(jié)

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