UVa #1025 A Spy in the Metro (例题9-1)

来源：互联网发布：华云数据面试编辑：程序博客网时间：2024/06/10 10:26

第一道动态规划。。书中给的状态转移的设计感觉很巧妙

要注意下数组大小避免RE。

Run Time: 0.032s

#define UVa  "LT9-1.1025.cpp"//A Spy in the Metrochar fileIn[30] = UVa, fileOut[30] = UVa;#include<cstring>#include<cstdio>#include<algorithm>using namespace std;//Global Variables. Reset upon Each Case!const int maxt = 4000 + 10, maxn = 50 + 10, maxm = 50 + 10, INF = 100000000;int N, T, M1, M2, t[maxn], d[maxm], e[maxm];int has_train[maxt][maxn][2];int dp[maxt][maxn];/////void init() {    memset(has_train, 0, sizeof(has_train));    scanf("%d", &T);    for(int i = 0; i < N-1; i ++) scanf("%d", &t[i]);    scanf("%d", &M1);    for(int i = 0; i < M1; i ++) scanf("%d", &d[i]);    scanf("%d", &M2);    for(int i = 0; i < M2; i ++) scanf("%d", &e[i]);    for(int i = 0; i < M1; i ++) {          //trains from left start point.        int time = d[i];        has_train[time][0][0] = 1;        for(int j = 0; j < N-1; j ++) {     //N-1 stations, N-2 intervals.            time += t[j];            has_train[time][j+1][0] = 1;        }    }    for(int i = 0; i < M2; i ++) {          //trains from right start point.        int time = e[i];        has_train[time][N-1][1] = 1;        for(int j = N-2; j >= 0; j --) {     //N-1 stations, N-2 intervals.            time += t[j];            has_train[time][j][1] = 1;        }    }}int main() {    int kase = 0;    while(scanf("%d", &N) && N) {        init();        for(int i = 0; i < N; i ++) dp[T][i] = INF;        for(int i = 0; i <= T; i ++) dp[i][N-1] = 0;        for(int i = T - 1; i >= 0; i --) {            for(int j = 0; j < N; j ++) {                dp[i][j] = dp[i+1][j] + 1;                if(has_train[i][j][0] && j < N-1 && i + t[j] <= T)                    dp[i][j] = min(dp[i][j], dp[i+t[j]][j+1]);                if(has_train[i][j][1] && j > 0 && 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][0] >= INF) printf("impossible\n");        else printf("%d\n", dp[0][0]);    }    return 0;}

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