UVA 590 - Always on the run

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第一次把题意理解错了，第二次因为一个 ] ，wrong了四次，还好在极值样例的检测下，找出来了。以后打代码一定要注意了。

我是这样定义状态的，f[i][j]表示在第i天时，在城市j所花费的最小费用，很简单的dp题：

代码如下：

#include<stdio.h>#include<string.h>#define INF 2000000000long long b[15][15][35];long long f[1010][15];int n, k, num;void output(){    if(f[k+1][n] < INF)printf("The best flight costs %lld.\n",f[k+1][n]);    else printf("No flight possible.\n");    printf("\n");}void solve(){    for(int i = 2; i <= n; i ++)    f[1][i] = INF;    f[1][1] = 0;    for(int i = 2; i <= k+1; i ++)        for(int j = 1; j <= n; j ++)        {            f[i][j] = INF;            for(int g = 1; g <= n; g ++)                if(g != j && f[i-1][g] != INF)                {                    if((i-1)%b[g][j][0]&&f[i-1][g]+b[g][j][(i-1)%b[g][j][0]] < f[i][j])                        f[i][j] = f[i-1][g] + b[g][j][(i-1)%b[g][j][0]];                    if(!((i-1)%b[g][j][0]) && f[i-1][g] + b[g][j][b[g][j][0]] < f[i][j])                        f[i][j] = f[i-1][g] + b[g][j][b[g][j][0]];                }    }    output();}void input(){    while(scanf("%d%d",&n, &k) == 2)    {        if(n==0&&k==0) break;        for(int i = 1; i <= n; i ++)            for(int j = 1; j <= n; j ++)                if(i != j)                {                    scanf("%d",&b[i][j][0]);                    for(int g = 1; g <= b[i][j][0]; g ++)                    {                        scanf("%d",&b[i][j][g]);                        if(b[i][j][g] == 0) b[i][j][g] = INF;                    }                }        printf("Scenario #%d\n",++num);        solve();    }}int main(){    num = 0;    input();    return 0;}

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