HDU 2282 Chocolate & HDU 2813 One fihgt one
2013-07-27 17:36
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两者都是求最小权匹配。
通过这次练习发现一个问题,就是求最小权匹配时,最好把W的赋值为-INF。在做第二题时,出现两者之间没有连边的匹配到了一块,所以最好赋值为-INF。输出时,只加上已经匹配边的长度。即:for(int i = 1; i <= m; i++) if(Left[i] && W[Left[i]][i] != -INF) ans += W[Left[i]][i];
通过这次练习发现一个问题,就是求最小权匹配时,最好把W的赋值为-INF。在做第二题时,出现两者之间没有连边的匹配到了一块,所以最好赋值为-INF。输出时,只加上已经匹配边的长度。即:for(int i = 1; i <= m; i++) if(Left[i] && W[Left[i]][i] != -INF) ans += W[Left[i]][i];
#include <iostream> #include <cstdlib> #include <cstdio> #include <string> #include <cstring> #include <cmath> #include <vector> #include <queue> #include <algorithm> #include <map> using namespace std; const int maxn = 510; const int INF = 0x3f3f3f3f; int n, m; int W[maxn][maxn]; int Lx[maxn], Ly[maxn]; int Left[maxn]; bool S[maxn], T[maxn]; bool match(int i) { S[i] = 1; for(int j = 1; j <= m; j++) if(Lx[i]+Ly[j] == W[i][j] && !T[j]) { T[j] = 1; if(!Left[j] || match(Left[j])) { Left[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 <= m; 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; } for(int j = 1; j <= m; j++) { if(T[j]) Ly[j] += a; } } void KM() { memset(Left, 0, sizeof(Left)); memset(Lx, 0, sizeof(Lx)); memset(Ly, 0, sizeof(Ly)); for(int i = 1; i <= n; i++) { for(int j = 1; j <= m; j++) Lx[i] = max(Lx[i], W[i][j]); } for(int i = 1; i <= n; i++) { for(;;) { for(int j = 1; j <= m; j++) S[j] = T[j] = 0; if(match(i)) break; else update(); } } } inline void readint(int &x) { char c; c = getchar(); while(!isdigit(c)) c = getchar(); x = 0; while(isdigit(c)) x = x*10+c-'0', c = getchar(); } inline void writeint(int x) { if(x > 9) writeint(x/10); putchar(x%10+'0'); } int num[maxn]; int N; int X[maxn], Y[maxn]; void read_case() { for(int i = 1; i <= N; i++) readint(num[i]); } void build() { n = m = 1; for(int i = 1; i <= N; i++) { if(!num[i]) Y[m++] = i; else { while(num[i] > 1) X[n++] = i, num[i] -= 1; } } n--, m--; for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) W[i][j] = -INF; for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) { W[i][j] = -min(abs(X[i]-Y[j]), N-(abs(X[i]-Y[j]))); } } void solve() { read_case(); build(); KM(); int ans = 0; for(int i = 1; i <= m; i++) if(Left[i] && W[Left[i]][i] != -INF) ans += W[Left[i]][i]; writeint(-ans), puts(""); } int main() { while(~scanf("%d", &N)) { solve(); } return 0; }
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