题意: 问是否存在长度为n的序列满足m个条件. 每个条件描述某一段子序列和大于或者小于k.
题解:差分约束的应用,将区间和维护成前缀和的差值,然后将各个位置上的前缀和用差分约束建立出来,用最长路求解,添加超级源点。SPFA中,如果一个点入队超过n次,那么表明此图存在正权值回路。
附上代码:
#include<bits/stdc++.h>
using namespace std;
const int inf=0x3f3f3f3f;
const int N=110;
struct edge{
int to,next,weight;
};
edge edges[4*N];
int head[N];
int n,m,tot;
int dis[N],num[N],vis[N];
void add_edge(int u,int v,int weight)
{
edges[tot].to=v;
edges[tot].weight=weight;
edges[tot].next=head[u];
head[u]=tot++;
}
int spfa(int u)
{
queue<int>q;
memset(dis,-inf,sizeof(dis));
memset(num,0,sizeof(num));
memset(vis,0,sizeof(vis));
q.push(u);
vis[u]=1;
dis[u]=0;
num[u]=1;
while(!q.empty()){
int x=q.front();
q.pop();
vis[x]=0;
for(int i=head[x];~i;i=edges[i].next){
int y=edges[i].to;
if(dis[y]<dis[x]+edges[i].weight){
dis[y]=dis[x]+edges[i].weight;
if(!vis[y]){
q.push(y);
vis[y]=1;
if(++num[y]>n){
return 0;
}
}
}
}
}
return 1;
}
int main()
{
while(~scanf("%d",&n)&&n){
tot=0;
memset(head,-1,sizeof(head));
scanf("%d",&m);
for(int i=0;i<m;i++){
int x,y,z;
char s[3];
scanf("%d%d%s%d",&x,&y,s,&z);
if(s[0]=='g'){
add_edge(x-1,x+y,z+1);
}else{
add_edge(x+y,x-1,-z+1);
}
}
for(int i=0;i<=n;i++){
add_edge(n+1,i,0);
}
if(spfa(n+1)){
printf("lamentable kingdom\n");
}else{
printf("successful conspiracy\n");
}
}
return 0;
}