Uva - 11383 - Golden Tiger Claw
题意:一个N*N的矩阵,第i行第j列的元素大小为w[i][j],每行求一个数row[i],每列求一个数col[j],使得row[i] + col[j] >= w[i][j],且所有的row[]与所有的col[]和总和最小( N <= 500, 其它输入数为正整数且 <= 100)。——>>row[i] + col[j] >= w[i][j],这个恰恰是二分图最佳完美匹配的一个式子,所以,以行row为X结点,以列col为Y结点,权值即为对应元素w[i][j]的值建图,跑一次KM就好。
另外发现:用scanf("%d", &N) == 1比用~scanf("%d", &N)快了3ms。。
#include <cstdio> #include <algorithm> using namespace std; const int maxn = 500 + 10; const int INF = 0x3f3f3f3f; int N, w[maxn][maxn], lx[maxn], ly[maxn], fa[maxn]; bool S[maxn], T[maxn]; bool match(int i){ S[i] = 1; for(int j = 1; j <= N; j++) if(lx[i] + ly[j] == w[i][j] && !T[j]){ T[j] = 1; if(!fa[j] || match(fa[j])){ fa[j] = i; return 1; } } return 0; } void update(){ int a = INF; for(int i = 1; i <= N; i++) if(S[i]) for(int j = 1; j <= N; j++) if(!T[j]) a = min(a, lx[i] + ly[j] - w[i][j]); for(int i = 1; i <= N; i++){ if(S[i]) lx[i] -= a; if(T[i]) ly[i] += a; } } void KM(){ for(int i = 1; i <= N; i++){ fa[i] = lx[i] = ly[i] = 0; for(int j = 1; j <= N; j++) lx[i] = max(lx[i], w[i][j]); } for(int i = 1; i <= N; i++) while(1){ for(int j = 1; j <= N; j++) S[j] = T[j] = 0; if(match(i)) break; else update(); } } void read(){ for(int i = 1; i <= N; i++) for(int j = 1; j <= N; j++) scanf("%d", &w[i][j]); } void solve(){ for(int i = 1; i < N; i++) printf("%d ", lx[i]); printf("%d\n", lx[N]); for(int i = 1; i < N; i++) printf("%d ", ly[i]); printf("%d\n", ly[N]); int sum = 0; for(int i = 1; i <= N; i++) sum += lx[i] + ly[i]; printf("%d\n", sum); } int main() { while(scanf("%d", &N) == 1){ read(); KM(); solve(); } return 0; }
补充:软件开发 , C++ ,