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先对于每种可能的PK情况进行判断胜负,然后求最小改动最大匹配。
#include <cstdlib> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <fstream> #include <iostream> #include <cctype> #define rep(i, l, r) for(int i=l; i<=r; i++) #define clr(x, c) memset(x, c, sizeof(x)) #define MAX 0x3fffffff #define N 100 using namespace std; int read() { int x=0; char ch=getchar(); while (!isdigit(ch)) ch=getchar(); while (isdigit(ch)) { x=x*10+ch-‘0‘; ch=getchar(); } return x; } int n, l[N], st[N], lx[N], ly[N], v[N][N], a1[N], a2[N], hp1[N], hp2[N], p[N]; bool vx[N], vy[N]; inline int pk(int x, int y) { int c=min((hp1[x]-1)/a2[y], (hp2[y]-1)/a1[x])+1; if (c*a1[x]>=hp2[y]) return p[x]; else return -p[x]; } bool Find(int x) { vx[x]=1; rep(y, 1, n) if (!vy[y]) { int a=lx[x]+ly[y]-v[x][y]; if (!a) { vy[y]=1; if (!l[y] || Find(l[y])) { l[y]=x; return 1; } } else st[y]=min(st[y], a); } return 0; } inline int km() { clr(ly, 0); clr(l, 0); rep(i, 1, n) lx[i]=-MAX; rep(i, 1, n) rep(j, 1, n) if (lx[i]<v[i][j]) lx[i]=v[i][j]; rep(i, 1, n) { rep(j, 1, n) st[j]=MAX; while (1) { clr(vx, 0); clr(vy, 0); if (Find(i)) break; int a=MAX; rep(j, 1, n) if (!vy[j] && a>st[j]) a=st[j]; rep(j, 1, n) if (vx[j]) lx[j]-=a; rep(j, 1, n) if (vy[j]) ly[j]+=a; else st[j]-=a; } } int a=0; rep(i, 1, n) a+=lx[i]+ly[i]; return a; } int main() { while (~scanf("%d", &n) && n) { rep(i, 1, n) p[i]=read(); rep(i, 1, n) hp1[i]=read(); rep(i, 1, n) hp2[i]=read(); rep(i, 1, n) a1[i]=read(); rep(i, 1, n) a2[i]=read(); rep(i, 1, n) rep(j, 1, n) v[i][j]=pk(i, j); rep(i, 1, n) rep(j, 1, n) v[i][j]*=N, v[i][j]+=i==j?1:0; int ans=km(); double k=ans%N; if (ans>0) printf("%d %.3lf%%\n", ans/N, k/n*100); else printf("Oh, I lose my dear seaco!\n"); } return 0; }
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原文地址:http://www.cnblogs.com/NanoApe/p/4382109.html