A Spy in the Metro UVA

动态规划，紫书上面已经解释很清楚了，逆向推理，从最后一个状态顺推前面全部状态，最终得到dp[0][0]，再进行一轮判断即可，具体实现见如下代码：

#include<iostream>#include<vector>#include<string>#include<set>#include<stack>#include<queue>#include<map>#include<algorithm>#include<cmath>#include<iomanip>#include<cstring>#include<sstream>#include<cstdio>#include<deque>using namespace std;int N, T;int train[500][500][2];int timeInter[500];int M1, M2;int toLeft[60],toRight[60];int dp[500][500];int main(){int Case = 1;while (cin >> N){if (N == 0) break;memset(train,0,sizeof(train));cin >> T;for (int i = 0; i < N-1; i++) cin >> timeInter[i];cin >> M1;for (int i = 0; i < M1; i++) cin >> toRight[i];cin >> M2;for (int i = 0; i < M2; i++) cin >> toLeft[i];for (int i = 0; i < M1; i++){int t = toRight[i];train[t][0][1] = 1;//时刻 车站编号  1 右for (int j = 0; j < N-1; j++){t += timeInter[j];train[t][j+1][1] = 1;}}for (int i = 0; i < M2; i++){int t = toLeft[i];train[t][N - 1][0] = 1;//0 左for (int j = N - 2; j >= 0; j--){t += timeInter[j];train[t][j][0] = 1;}}memset(dp,0,sizeof(dp));dp[T][N - 1] = 0;for (int i = 0; i < N - 1; i++) dp[T][i] = 1 << 20;for (int i = T - 1; i >= 0; i--){for (int j = 0; j < N; j++){dp[i][j] = dp[i+1][j] + 1;if (j < N - 1 && train[i][j][1] && i + timeInter[j] <= T){//rightdp[i][j] = min(dp[i][j],dp[i+timeInter[j]][j+1]);}if (j > 0 && train[i][j][0] && i + timeInter[j-1] <= T){//leftdp[i][j] = min(dp[i][j], dp[i + timeInter[j - 1]][j-1]);}}}cout << "Case Number "<<Case++<<": ";if (dp[0][0] >= (1 << 20)) cout << "impossible" << endl;else cout << dp[0][0] << endl;}return 0;}