pku 1015 Jury Compromise DP

#include <iostream> #include <cstdio> #include <cstring> #include <exception> using namespace std; #define BASE 400 int opt[202][22][802]; int arr1[202], arr2[202]; int ret[22]; void findPath(int n,int m, int delta) { int i = n, j = m; while(i >= 1 && j >= 1) { if(opt[i-1][j][BASE+delta] == opt[i][j][BASE+delta]) i--; else { delta -= arr1[i]-arr2[i]; ret[j] = i; i--, j--; } } } int main() { int n, m; int cnt = 0; while (scanf("%d%d", &n, &m) && n) { for(int i = 1; i <= n; ++i) scanf("%d%d", &arr1[i], &arr2[i]); int scope = m*20; memset(opt, -1, sizeof(opt)); for(int i = 1; i <= n; ++i) opt[i][0][BASE] = 0; opt[1][1][BASE+arr1[1]-arr2[1]] = arr1[1]+arr2[1]; for(int i = 2; i <= n; ++i) for(int j = 1; j <= m; ++j) for (int k = -scope; k <= scope; ++k) { if(j <= i) { if(j <= i-1) { int v = BASE+k-(arr1[i]-arr2[i]); if(v >= BASE-scope && v <= BASE+scope) { if(opt[i-1][j-1][v] != -1) opt[i][j][BASE+k] = max(opt[i][j][BASE+k], opt[i-1][j-1][v]+arr1[i]+arr2[i]); if(opt[i-1][j][BASE+k] != -1) opt[i][j][BASE+k] = max(opt[i][j][BASE+k], opt[i-1][j][BASE+k]); } else { if(opt[i-1][j][BASE+k] != -1) opt[i][j][BASE+k] = max(opt[i][j][BASE+k], opt[i-1][j][BASE+k]); } } else { int v = BASE+k-(arr1[i]-arr2[i]); if(v >= BASE-scope && v <= BASE+scope) { if(opt[i-1][j-1][v] != -1) opt[i][j][BASE+k] = max(opt[i][j][BASE+k], opt[i-1][j-1][v]+arr1[i]+arr2[i]); } } } } int delta = 0; for(int i = 0; i <= scope; ++i) { if(opt[n][m][BASE+i] != -1) { if(opt[n][m][BASE-i] == -1) delta = i; else { if(opt[n][m][BASE+i] > opt[n][m][BASE-i]) delta = i; else delta = -i; } break; } else if(opt[n][m][BASE-i] != -1) { delta = -i; break; } } findPath(n, m, delta); int sum1 = 0, sum2 = 0; for(int i = 1; i <= m; ++i) sum1 += arr1[ret[i]], sum2 += arr2[ret[i]]; printf("Jury #%d/nBest jury has value %d for prosecution and value %d for defence:/n", ++cnt, sum1, sum2); for(int i = 1; i <= m; ++i) printf(" %d", ret[i]); printf("/n/n"); } return 0; } 

 

下面附一份我AC之前的一份丑陋的WA代码, 开始没有考虑全面, 状态有后效应, 需增加状态转移的维数来消除后效应.

 

//WA代码!! #include <iostream> #include <cmath> using namespace std; int delta[205][25]; int sum1[205][25], sum2[205][25]; bool record[205][25]; int arr1[205],arr2[205]; int ret[205]; int i, j; int main() { int n, m; int cnt = 0; while(scanf("%d%d", &n, &m) && n) { for(i = 1; i <= n; ++i) scanf("%d%d", &arr1[i], &arr2[i]); delta[1][0] = 0; sum1[1][0] = 0, sum2[1][0] = 0; record[1][0] = false; delta[1][1] = arr1[1]-arr2[1]; sum1[1][1] = arr1[1], sum2[1][1] = arr2[1]; record[1][1] = true; for(i = 2; i <= n; ++i) for(j = 0; j <= m; ++j) if(j <= i) { if(j <= i-1) { if(abs(delta[i-1][j]) > abs(delta[i-1][j-1]+arr1[i]-arr2[i])) { delta[i][j] = delta[i-1][j-1]+arr1[i]-arr2[i]; sum1[i][j] = sum1[i-1][j-1]+arr1[i], sum2[i][j] = sum2[i-1][j-1]+arr2[i]; record[i][j] = true; } else if(abs(delta[i-1][j]) < abs(delta[i-1][j-1]+arr1[i]-arr2[i])) { delta[i][j] = delta[i-1][j]; sum1[i][j] = sum1[i-1][j], sum2[i][j] = sum2[i-1][j]; record[i][j] = false; } else { if(sum1[i-1][j]+sum2[i-1][j] <= sum1[i-1][j-1]+arr1[i]+sum2[i-1][j-1]+arr2[i]) { delta[i][j] = delta[i-1][j-1]+arr1[i]-arr2[i]; sum1[i][j] = sum1[i-1][j-1]+arr1[i], sum2[i][j] = sum2[i-1][j-1]+arr2[i]; record[i][j] = true; } else { delta[i][j] = delta[i-1][j]; sum1[i][j] = sum1[i-1][j], sum2[i][j] = sum2[i-1][j]; record[i][j] = false; } } } else { delta[i][j] = delta[i-1][j-1]+arr1[i]-arr2[i]; sum1[i][j] = sum1[i-1][j-1]+arr1[i], sum2[i][j] = sum2[i-1][j-1]+arr2[i]; record[i][j] = true; } } cout << "delta: " << delta[n][m] << endl; i = n; j = m; int num = 1; while(i >= 1 && j >= 0) { if(record[i][j]) { ret[num++] = i; i--, j--; } else i--; } int sum1 = 0, sum2 = 0; for(i = 1; i <= m; ++i) { sum1 += arr1[ret[i]]; sum2 += arr2[ret[i]]; } printf("Jury #%d/nBest jury has value %d for prosecution and value %d for defence:/n", ++cnt, sum1, sum2); for(i = m; i >= 1; --i) printf(" %d", ret[i]); printf("/n/n"); } return 0; }