uvalive4256(DP)

来源：互联网发布：华为手机数据迁移编辑：程序博客网时间：2024/04/30 04:51

题意：

给出一张图，然后给出一个序列，修改序列中一些数字，要求使这个序列相邻的两个点．要么是相同的点，要么在图中是相邻点；

思路：

dp[i][j]代表序列前ｉ个，并且第ｉ个的值是ｊ　，满足这一关系，最少修改几次几次；

那么dp[i][j] ＝ min ( dp[i - 1][k])　（ｋ与ｊ是相连的）；

#include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace std;const int N = 105;int INF = 0x3f3f3f3f;int n1,n2,l;int suq[2 * N];int dp[2 * N][N];vector<int> v[N];int main(){int t;scanf("%d",&t);while(t--) {memset(dp, 0 ,sizeof(dp));scanf("%d%d",&n1,&n2);for(int i = 0; i <= n1; i++) {v[i].clear();v[i].push_back(i);}int x,y;for(int i = 0; i < n2; i++) {scanf("%d%d",&x,&y);v[x].push_back(y);v[y].push_back(x);}scanf("%d",&l);for(int i = 1; i <= l; i++) {scanf("%d",&suq[i]);}for(int i = 1; i <= l; i++) {for(int j = 1; j <= n1; j++) {dp[i][j] = INF;for(int k = 0; k < v[j].size(); k++) {if(j == suq[i])dp[i][j] = min(dp[i][j], dp[i - 1][v[j][k]]);elsedp[i][j] = min(dp[i][j], dp[i - 1][v[j][k]] + 1);}}}int m = INF;for(int i = 1; i <= n1; i++) {m = min(m, dp[l][i]);}printf("%d\n",m);}return 0;}

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