A - Brainwashing Device
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Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status Practice URAL 1900
Description
While some people travel in space from planet to planet and discover new worlds, the others who live on Earth still have to get up in the morning, go to work, return back home and try to have a rest. They don't like this situation anymore and envy people who can afford space travel.
That doesn't suit the government of the Earth. Their goal is to make everyone happy, but that is difficult. That is why the government decided to do some tests in the small town of Lux, and then, if everything goes well, to apply the experience to other cities.
Lux's distinctive feature is that it is situated in a long underground tunnel and there is only one train inside it. Almost everyone in the town uses the train. The government has bought a brainwashing device and installed it in the train. The device is new and its effects aren't well understood yet, so the government decided to limit the number of the spans where the device would be turned on. Statistics on number of passengers travelling between each pair of stations every morning have already been collected. Now the government should pick the spans where the device could be used to make most of the people happy.
Input
The first line contains integers n and k that are the total number of the stations and the number of spans between adjacent stations where the device could be turned on (2 ≤ n ≤ 500; 1 ≤ k ≤ n − 1). The i'th of the next n − 1 lines contains n − i integers. The j'th integer is the number of passengers traveling from i'th to ( i + j)'th station. These numbers are non-negative and don't exceed 100. You can assume that every passenger uses the train only once a day.
Output
In the first line output the total number of people who will become happy because of the device. In the second line output k integers in the ascending order that are the indexes of the spans where the device should be turned on. The span between the station i and i + 1 has the index i. If the problem has multiple solutions you can output any of them.
Sample Input
input
output
4 15 0 65 35
143
思路：dp[i][K]前i个选了K个的最优值，i必选，并不是一道很难的题，预处理出dis[i][j]放i所得的值再放j所重复的值。
坑：全部都要放上，而且要升序
#include<iostream>#include<cstring>#include<cstdio>#define FOR(i,a,b) for(int i=a;i<=b;++i)#define clr(f,z) memset(f,z,sizeof(f))using namespace std;const int oo=0x3f3f3f3f;const int mm=550;int g[mm][mm],dis[mm][mm],c[mm][mm];int f[mm],dp[mm][mm],pre[mm][mm],res[mm];bool vis[mm];int n,K;int lowbit(int x){ return x&(-x);}void add(int x,int y,int z){ for(int i=x;i<mm;i+=lowbit(i)) for(int j=y;j<mm;j+=lowbit(j)) c[i][j]+=z;}int query(int x,int y){ int ret=0; for(int i=x;i>0;i-=lowbit(i)) for(int j=y;j>0;j-=lowbit(j)) ret+=c[i][j]; return ret;}void print(int x,int k){ vis[x]=1; if(pre[x][k]!=-1) {print(pre[x][k],k-1); //printf(" %d",x); } else //printf("%d",x); { --k; FOR(i,1,n) if(vis[i]) printf("%d ",i); else if(k) printf("%d ",i),k--; }}int main(){ int x; while(~scanf("%d%d",&n,&K)) { clr(c,0); FOR(i,1,n-1) FOR(j,i,n-1) {scanf("%d",&x); add(i,j,x); } --n; FOR(i,1,n) FOR(j,i,n) dis[i][j]=query(i,n)-query(i,j-1); clr(dp,0); clr(pre,-1); FOR(i,1,n)dp[i][1]=dis[i][i];///放在i上 FOR(i,2,K) FOR(j,i,n)//j点之前选i点的最优 { dp[j][i]=dp[j][i-1]; FOR(k,1,j-1) if(dp[j][i]<dp[k][i-1]+dis[j][j]-dis[k][j]) dp[j][i]=dp[k][i-1]+dis[j][j]-dis[k][j],pre[j][i]=k; // printf("zd=%d %d %d\n",j,i,dp[j][i]); } clr(vis,0); int ans=-1,id; FOR(i,1,n) if(dp[i][K]>ans) ans=dp[i][K],id=i; printf("%d\n",ans); print(id,K); printf("\n"); }}