USACO 3.3.5

最新推荐文章于 2022-12-02 22:05:12 发布

原创最新推荐文章于 2022-12-02 22:05:12 发布 · 417 阅读

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博弈问题，可以使用dp解决.
f[i][j] = s[i][j] - min{f[i+1][j], f[i][j-1]}
s[i][j]表示从i加到j的和, 枚举l=j-i.

Executing...
   Test 1: TEST OK [0.011 secs, 3108 KB]
   Test 2: TEST OK [0.011 secs, 3108 KB]
   Test 3: TEST OK [0.011 secs, 3108 KB]
   Test 4: TEST OK [0.011 secs, 3108 KB]
   Test 5: TEST OK [0.011 secs, 3108 KB]
   Test 6: TEST OK [0.011 secs, 3108 KB]
   Test 7: TEST OK [0.000 secs, 3108 KB]
   Test 8: TEST OK [0.000 secs, 3108 KB]
   Test 9: TEST OK [0.000 secs, 3108 KB]
   Test 10: TEST OK [0.000 secs, 3108 KB]
   Test 11: TEST OK [0.011 secs, 3108 KB]
   Test 12: TEST OK [0.011 secs, 3108 KB]
   Test 13: TEST OK [0.000 secs, 3108 KB]
   Test 14: TEST OK [0.000 secs, 3108 KB]
   Test 15: TEST OK [0.011 secs, 3108 KB]
   Test 16: TEST OK [0.000 secs, 3108 KB]

All tests OK.

YOUR PROGRAM ('game1') WORKED FIRST TIME! That's fantastic -- and a rare thing. Please accept these special automated congratulations.

[Code]


#include<stdio.h>
#include<iostream>
using namespace std;
const int MAXN = 100+10;
int N[MAXN], f[MAXN][MAXN] = {0}, s[MAXN][MAXN] = {0}, sum[MAXN] = {0};
int min(int x, int y) {return (x < y) ? x : y;}
int main()
{
    FILE *fin, *fout;
    fin    = fopen("game1.in", "r");
    fout = fopen("game1.out", "w");
    int n=0, i, j, l;
    fscanf(fin, "%d", &n);
    for (i = 1; i <= n; i++) fscanf(fin, "%d", &N[i]);
    for (i = 1; i <= n; i++)
    {
        s[i][i] = N[i];
        for (j = i+1; j <= n; j++)
            s[i][j] = s[i][j-1] + N[j];
    }
    for (i = 1; i <= n; i++) f[i][i] = N[i];
    for (l = 1; l < n; l++)
        for (i = 1; i <= n-l; i++)
            f[i][i+l] = s[i][i+l] - min(f[i+1][i+l], f[i][i+l-1]);
    fprintf(fout, "%d %d/n", f[1][n], s[1][n]-f[1][n]);
}
 