UVa 1025

动态规划基础例题，要理解状态和状态转移方程。。。

#include <bits/stdc++.h>
using namespace std;

int kase = 0;
int n, T, t[80];
int M1, M2, st1[64], st2[64];
bool has_train[512][64][2];
int dp[512][80];
const int INF = 1<<8;


int main() {
    //freopen("in.txt", "r", stdin);
    while(cin >> n && n) {
        memset(dp, 0, sizeof(dp));
        memset(has_train, false, sizeof(has_train));
        memset(st1, 0, sizeof(st1));
        memset(st2, 0, sizeof(st2));

        cin >> T;
        for(int i = 1; i < n; ++i) cin >> t[i];
        cin >> M1;
        for(int i = 1; i <= M1; ++i) {
            cin >> st1[i];
            int time = st1[i];
            for(int j = 1; j <= n; ++j) {
                has_train[time][j][0] = true;
                time += t[j];
                if(time > 202) break;
            }
        }
        cin >> M2;
        for(int i = 1; i <= M2; ++i) {
            cin >> st2[i];
            int time = st2[i];
            for(int j = n; j >= 1; --j) {
                has_train[time][j][1] = true;
                time += t[j-1];
                if(time > 202) break;
            }
        }

        for(int i = 1; i < n; ++i) dp[T][i] = INF;
        dp[T][n] = 0;
        for(int i = T-1; i >= 0; --i) {
            for(int j = 1; j <= n; ++j) {
                dp[i][j] = dp[i+1][j] + 1;
                if(j < n && has_train[i][j][0] && i+t[j] <= T)
                    dp[i][j] = min(dp[i][j], dp[i+t[j]][j+1]);
                if(j > 1 && has_train[i][j][1] && i+t[j-1] <= T)
                    dp[i][j] = min(dp[i][j], dp[i+t[j-1]][j-1]);
            }
        }
        printf("Case Number %d: ", ++kase);
        if(dp[0][1] >= INF) cout << "impossible" << endl;
        else cout << dp[0][1] << endl;
    }
    return 0;
}