UVALive - 4256 Salesmen DP+暴力

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题目大意：给定一个包含n1个点，n2条边的无向连通图，再给出一个序列长度为n的整数序列A，要求每两个相邻的整数不是相等就是连通，给出的这个序列上的数可以修改，问至少经过多少次的修改可以得到正确的序列

解题思路：暴力枚举每种状态，设dp[i][j]为第i个以j结尾时共修改了多少次，这样就可以得到状态转义方程了

dp[i][j] = min(dp[i][j],dp[i-1][k]+1)这是修改了第i个数的情况

dp[i][j] = min(dp[i][j],dp[i-1][k])这是没有修改第i个数的情况

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;#define maxn1 110#define maxn2 210#define INF 0x3f3f3f3fint n1, n2, n,  graph[maxn1][maxn1], A[maxn2], dp[maxn2][maxn2];int main() {int test;scanf("%d", &test); while(test--) {scanf("%d%d", &n1, &n2);memset(graph,0,sizeof(graph));int x, y;for(int i = 0; i < n2; i++) {scanf("%d%d",&x, &y);graph[x][y] = graph[y][x] = 1;}for(int i = 1; i <= n1; i++) graph[i][i] = 1;scanf("%d",&n);for(int i = 1; i <= n; i++) scanf("%d", &A[i]);for(int i = 1; i <= n1; i++)dp[0][i] = 0;for(int i = 1; i <= n; i++)for(int j = 1; j <= n1; j++) {dp[i][j] = INF;for(int k = 1; k <= n1; k++) if(j == k || graph[j][k]) {if(j == A[i])dp[i][j] = min(dp[i][j],dp[i-1][k]);elsedp[i][j] = min(dp[i][j],dp[i-1][k]+1);}}int ans = dp[n][1];for(int i = 1; i <= n1; i++)ans = min(ans,dp[n][i]);printf("%d\n",ans);}return 0;}

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