P1772 [ZJOI2006]物流运输 最短路+DP

思路：最短路+DP

提交：1次

题解：

$f[i]$表示到第$i$天的最小代价，我们可以预先处理出$i,j$两天之间(包括$i,j$)都可通行的最短路的代价记做$s[i][j]$，然后有$f[i]=min(f[i],f[j]+s[j+1][i]*(i-j)+W);$

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#define ull unsigned long long
#define ll long long
#define R register int
using namespace std;
#define pause (for(R i=1;i<=10000000000;++i))
#define In freopen("NOIPAK++.in","r",stdin)
#define Out freopen("out.out","w",stdout)
namespace Fread {
static char B[<<],*S=B,*D=B;
#ifndef JACK
#define getchar() (S==D&&(D=(S=B)+fread(B,1,1<<15,stdin),S==D)?EOF:*S++)
#endif
inline int g() {
    R ret=,fix=; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-:fix;
    if(ch==EOF) return EOF; do ret=ret*+(ch^); while(isdigit(ch=getchar())); return ret*fix;
} inline bool isempty(const char& ch) {return (ch<=||ch>=);}
inline void gs(char* s) {
    register char ch; while(isempty(ch=getchar()));
    do *s++=ch; while(!isempty(ch=getchar()));
}
} using Fread::g; using Fread::gs;

namespace Luitaryi {
const int N=,M=;
int t,n,W,m,cnt;
int vr[N*N],nxt[N*N],w[N*N],fir[N],d[N],s[M][M];
ll f[N];
bool vis[N],ban[N],c[N][M];
inline void add(int u,int v,int ww) {vr[++cnt]=v,w[cnt]=ww,nxt[cnt]=fir[u],fir[u]=cnt;}
inline int spfa() {
    memset(d,0x3f,sizeof(d)); memset(vis,,sizeof(vis));
    queue<int> q; q.push(); d[]=; vis[]=true;
    while(q.size()) {
        R u=q.front(); q.pop(); vis[u]=false;
        if(ban[u]) continue;
        for(R i=fir[u];i;i=nxt[i]) { R v=vr[i];
            if(d[v]>d[u]+w[i]) {
                d[v]=d[u]+w[i];
                if(!vis[v]) q.push(v),vis[v]=true;
            }
        }
    } return d[n];
}
inline void main() {
    t=g(),n=g(),W=g(),m=g();
    for(R i=,u,v,w;i<=m;++i) u=g(),v=g(),w=g(),add(u,v,w),add(v,u,w);
    m=g(); for(R i=,u,l,r;i<=m;++i) {
        u=g(),l=g(),r=g(); for(R j=l;j<=r;++j) c[u][j]=true;
    } for(R i=;i<=t;++i) { memset(ban,,sizeof(ban));
        for(R j=i;j<=t;++j) {
            for(R u=;u<=n;++u) ban[u]|=c[u][j];
            s[i][j]=spfa();
        }
    }
    for(R i=;i<=t;++i) { f[i]=1ll*s[][i]*i;
        for(R j=;j<i;++j) f[i]=min(f[i],f[j]+1ll*s[j+][i]*(i-j)+W);
    } printf("%lld\n",f[t]);
}
}
signed main() {
    Luitaryi::main();
}

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2019.07.21