ZOJ 2297 Survival 狀態壓縮DP

做法：這種元素個數很少，而且狀態和順序有關，可以往狀態壓縮上考慮

#include<cstdio>#include<cstring>const int LMT=1<<20;int dp[LMT],c[25],r[25];int max(int a,int b){return a>b?a:b;}int min(int a,int b){return a<b?a:b;}void init(void){memset(dp,-1,sizeof(dp));}int main(void){int n,i,j,lim,boss;while(~scanf("%d",&n)){init();for(i=1;i<n;i++)scanf("%d%d",&c[i],&r[i]);scanf("%d",&boss);lim=1<<(n-1);dp[0]=100;for(i=0;i<lim;i++)if(dp[i]!=-1){for(j=1;j<n;j++)if(!(i&(1<<(j-1)))&&dp[i]>=c[j]){   dp[i|(1<<(j-1))]=max(dp[i|(1<<(j-1))],dp[i]+r[j]-c[j]);   dp[i|(1<<(j-1))]=min(dp[i|(1<<(j-1))],100);}}if(dp[lim-1]<boss||dp[lim-1]==-1)printf("try again\n");else printf("clear!!!\n");}return 0;}