OJ-Problems-Source/HDOJ/3236_stay_accept.cpp

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#include <vector>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
using namespace std;
int dp[505][55][2];
struct node{
int v,w;
};
int main(){
int V1,V2,i,j,a,b,c,k,n,ans,cas,sum,pre,sign;
cas=1; //二维带限制条件的01背包
while(scanf("%d%d%d",&V1,&V2,&n)!=EOF&&(V1||V2||n)){
vector<node> G[2]; //dp[i][j][0]代表容量是i和j没有用免费机会时的最大值
memset(dp,-1,sizeof(dp)); //dp[i][j][0]代表容量是i和j用免费机会时的最大值
for(i=0;i<n;i++){
scanf("%d%d%d",&a,&b,&c);
G[c].push_back((node){a,b});
}
sum=dp[0][0][0]=0;
for(i=0;i<G[1].size();i++){
pre=sum; //只有前一个的值是pre才能
sum+=G[1][i].w; //继续往里装
for(j=V1;j>=0;j--){
for(k=V2;k>=0;k--){
if(dp[j][k][0]==pre)
dp[j][k][1]=sum;
if(j>=G[1][i].v){
if(dp[j-G[1][i].v][k][1]==pre)
dp[j][k][1]=sum;
if(dp[j-G[1][i].v][k][0]==pre)
dp[j][k][0]=sum;
}
if(k>=G[1][i].v){
if(dp[j][k-G[1][i].v][1]==pre)
dp[j][k][1]=sum;
if(dp[j][k-G[1][i].v][0]==pre)
dp[j][k][0]=sum;
}
}
}
}
sign=0;
for(i=0;i<=V1;i++)
for(j=0;j<=V2;j++)
for(k=0;k<=1;k++){
if(dp[i][j][k]!=sum)
dp[i][j][k]=-1;
else
sign=1;
}
if(!sign){
printf("Case %d: -1\n\n",cas++);
continue;
} //看是否能装满特殊物品
for(i=0;i<G[0].size();i++){
for(j=V1;j>=0;j--){
for(k=V2;k>=0;k--){
if(dp[j][k][0]!=-1) //要放在最前面,避免了重复装入
dp[j][k][1]=max(dp[j][k][1],dp[j][k][0]+G[0][i].w);
if(j>=G[0][i].v){
if(dp[j-G[0][i].v][k][1]!=-1)
dp[j][k][1]=max(dp[j][k][1],dp[j-G[0][i].v][k][1]+G[0][i].w);
if(dp[j-G[0][i].v][k][0]!=-1)
dp[j][k][0]=max(dp[j][k][0],dp[j-G[0][i].v][k][0]+G[0][i].w);
}
if(k>=G[0][i].v){
if(dp[j][k-G[0][i].v][1]!=-1)
dp[j][k][1]=max(dp[j][k][1],dp[j][k-G[0][i].v][1]+G[0][i].w);
if(dp[j][k-G[0][i].v][0]!=-1)
dp[j][k][0]=max(dp[j][k][0],dp[j][k-G[0][i].v][0]+G[0][i].w);
}
}
}
}
ans=0;
for(i=0;i<=V1;i++)
for(j=0;j<=V2;j++)
for(k=0;k<=1;k++)
ans=max(ans,dp[i][j][k]);
printf("Case %d: %d\n\n",cas++,ans);
}
return 0;
}